WEBVTT 00:00:00.000 --> 00:00:00.960 align:middle line:90% [SQUEAKING] 00:00:00.960 --> 00:00:03.360 align:middle line:90% [RUSTLING] 00:00:03.360 --> 00:00:05.280 align:middle line:90% [CLICKING] 00:00:05.280 --> 00:00:10.080 align:middle line:90% 00:00:10.080 --> 00:00:11.060 align:middle line:90% [SIDE CONVERSATION] 00:00:11.060 --> 00:00:12.390 align:middle line:90% ZACHARY ABEL: Good morning. 00:00:12.390 --> 00:00:13.350 align:middle line:90% Good afternoon. 00:00:13.350 --> 00:00:15.140 align:middle line:90% Let's go ahead and get started. 00:00:15.140 --> 00:00:17.070 align:middle line:90% Welcome to lecture 16. 00:00:17.070 --> 00:00:19.250 align:middle line:84% This is the last lecture worth of material 00:00:19.250 --> 00:00:21.650 align:middle line:84% that we're covering before our quiz. 00:00:21.650 --> 00:00:22.500 align:middle line:90% All right. 00:00:22.500 --> 00:00:24.000 align:middle line:90% Let's talk about counting. 00:00:24.000 --> 00:00:28.370 align:middle line:90% This is lecture 16 on counting. 00:00:28.370 --> 00:00:30.350 align:middle line:84% And where we left off last time was 00:00:30.350 --> 00:00:33.900 align:middle line:84% I had just mentioned the generalized product rule. 00:00:33.900 --> 00:00:36.530 align:middle line:84% And we used the generalized product rule 00:00:36.530 --> 00:00:39.440 align:middle line:84% to count the number of permutations 00:00:39.440 --> 00:00:43.790 align:middle line:84% of a deck of comically oversized playing cards. 00:00:43.790 --> 00:00:45.480 align:middle line:90% We had 52 cards. 00:00:45.480 --> 00:00:47.870 align:middle line:84% And we decided if we want to choose an order among all 00:00:47.870 --> 00:00:51.920 align:middle line:84% these cards, well, we have 52 options for the first card. 00:00:51.920 --> 00:00:53.660 align:middle line:90% Queen of diamonds this time. 00:00:53.660 --> 00:00:57.870 align:middle line:84% After that, we have 51 options left for the second card. 00:00:57.870 --> 00:01:00.030 align:middle line:90% Four of clubs this time. 00:01:00.030 --> 00:01:03.390 align:middle line:84% Now we have 50 cards left, 50 options. 00:01:03.390 --> 00:01:10.080 align:middle line:84% And notice whatever first two cards I chose initially, 00:01:10.080 --> 00:01:12.510 align:middle line:84% whatever cards I chose is going to change the set 00:01:12.510 --> 00:01:14.800 align:middle line:90% of available options left. 00:01:14.800 --> 00:01:17.050 align:middle line:84% Instead of choosing the queen of diamonds first, 00:01:17.050 --> 00:01:21.130 align:middle line:84% if I had chosen the eight of clubs first, 00:01:21.130 --> 00:01:22.770 align:middle line:84% then it would be this 50 cards that 00:01:22.770 --> 00:01:27.120 align:middle line:84% are remaining on step three rather than the previous set. 00:01:27.120 --> 00:01:31.530 align:middle line:84% But still, on step three, we always have 50 choices left. 00:01:31.530 --> 00:01:34.150 align:middle line:84% So now we're choosing the two of hearts. 00:01:34.150 --> 00:01:36.840 align:middle line:84% On step four, we always have 49 choices 00:01:36.840 --> 00:01:40.770 align:middle line:90% left, then 48, 47, et cetera. 00:01:40.770 --> 00:01:45.990 align:middle line:84% And so we were able to sum this up into a general principle 00:01:45.990 --> 00:01:47.740 align:middle line:84% called the generalized product rule, 00:01:47.740 --> 00:01:51.720 align:middle line:84% which basically says if you're making a sequence of choices, 00:01:51.720 --> 00:01:55.560 align:middle line:84% and at the i-th choice, you always 00:01:55.560 --> 00:02:00.120 align:middle line:84% have a fixed number, ni, of possible choices 00:02:00.120 --> 00:02:03.540 align:middle line:84% at that i-th step, then the number of ways 00:02:03.540 --> 00:02:05.100 align:middle line:84% to make all your choices in sequence 00:02:05.100 --> 00:02:09.270 align:middle line:84% is just n1 times n2 times n3 all the way to nk. 00:02:09.270 --> 00:02:12.070 align:middle line:84% And that's what this says in text. 00:02:12.070 --> 00:02:16.140 align:middle line:84% So if A is a set of length k sequences where there 00:02:16.140 --> 00:02:19.480 align:middle line:84% are exactly n1 possible choices for the first one, 00:02:19.480 --> 00:02:21.540 align:middle line:84% so 52 choices for the first card, 00:02:21.540 --> 00:02:25.000 align:middle line:84% and then n2 possible choices at the second step, 00:02:25.000 --> 00:02:27.990 align:middle line:84% no matter what you chose for the first step, then 00:02:27.990 --> 00:02:30.660 align:middle line:84% n3 for the third step, no matter what 00:02:30.660 --> 00:02:33.640 align:middle line:84% your first two choices were, and so on all the way down. 00:02:33.640 --> 00:02:36.060 align:middle line:84% Then the size of your set is just 00:02:36.060 --> 00:02:38.100 align:middle line:84% the product of the number of choices 00:02:38.100 --> 00:02:40.170 align:middle line:84% you have in step one times the number of choices 00:02:40.170 --> 00:02:43.290 align:middle line:84% you have in step two, et cetera, et cetera. 00:02:43.290 --> 00:02:45.360 align:middle line:90% Does that idea make sense? 00:02:45.360 --> 00:02:48.690 align:middle line:90% Let's do one more example here. 00:02:48.690 --> 00:02:52.350 align:middle line:90% And this example is about money. 00:02:52.350 --> 00:02:55.260 align:middle line:90% Money is a fun topic, right? 00:02:55.260 --> 00:02:57.260 align:middle line:84% So anyone who has a dollar bill on them, 00:02:57.260 --> 00:02:59.400 align:middle line:84% doesn't have to be a one, just any dollar bill. 00:02:59.400 --> 00:03:02.100 align:middle line:84% Go ahead and pull it out for a second and then give it to me. 00:03:02.100 --> 00:03:03.680 align:middle line:90% No, you can keep it. 00:03:03.680 --> 00:03:05.700 align:middle line:84% But take a look at the serial number. 00:03:05.700 --> 00:03:09.720 align:middle line:84% So every dollar bill has a serial number, 00:03:09.720 --> 00:03:15.633 align:middle line:84% which is a length eight sequence of digits. 00:03:15.633 --> 00:03:18.710 align:middle line:90% 00:03:18.710 --> 00:03:21.138 align:middle line:84% How many possible serial numbers are there? 00:03:21.138 --> 00:03:22.680 align:middle line:84% There might be some letters in there. 00:03:22.680 --> 00:03:23.820 align:middle line:90% Ignore the letters. 00:03:23.820 --> 00:03:25.260 align:middle line:90% There are eight digits. 00:03:25.260 --> 00:03:28.860 align:middle line:84% How many possible serial numbers are there? 00:03:28.860 --> 00:03:30.260 align:middle line:84% How many ways are there to choose 00:03:30.260 --> 00:03:33.345 align:middle line:90% a sequence of eight digits? 00:03:33.345 --> 00:03:33.845 align:middle line:90% Yes? 00:03:33.845 --> 00:03:35.180 align:middle line:90% AUDIENCE: Eight factorial. 00:03:35.180 --> 00:03:37.160 align:middle line:90% ZACHARY ABEL: 8 factorial? 00:03:37.160 --> 00:03:40.610 align:middle line:90% Eight factorial serials. 00:03:40.610 --> 00:03:42.270 align:middle line:90% Any other suggestions? 00:03:42.270 --> 00:03:42.785 align:middle line:90% Yes. 00:03:42.785 --> 00:03:43.910 align:middle line:90% AUDIENCE: 10 to the eighth. 00:03:43.910 --> 00:03:45.230 align:middle line:90% ZACHARY ABEL: 10 to the eighth. 00:03:45.230 --> 00:03:47.390 align:middle line:90% 10 to the eighth. 00:03:47.390 --> 00:03:49.280 align:middle line:90% Any other suggestions? 00:03:49.280 --> 00:03:50.545 align:middle line:90% Yes? 00:03:50.545 --> 00:03:52.670 align:middle line:84% AUDIENCE: [INAUDIBLE] all the way down [INAUDIBLE]. 00:03:52.670 --> 00:03:53.480 align:middle line:90% ZACHARY ABEL: OK. 00:03:53.480 --> 00:03:57.770 align:middle line:84% 10 times 9 times 8, all the way down to 3, 00:03:57.770 --> 00:03:59.770 align:middle line:84% because we're only choosing eight digits instead 00:03:59.770 --> 00:04:02.350 align:middle line:90% of choosing all 10 digits. 00:04:02.350 --> 00:04:02.990 align:middle line:90% OK. 00:04:02.990 --> 00:04:05.410 align:middle line:84% I'm going to put another suggestion in there, 8 00:04:05.410 --> 00:04:06.120 align:middle line:90% to the 10. 00:04:06.120 --> 00:04:08.920 align:middle line:90% 00:04:08.920 --> 00:04:10.395 align:middle line:90% So which of these is right? 00:04:10.395 --> 00:04:13.180 align:middle line:90% 00:04:13.180 --> 00:04:15.580 align:middle line:84% So importantly, I never said that the digits 00:04:15.580 --> 00:04:16.880 align:middle line:90% had to be distinct. 00:04:16.880 --> 00:04:20.170 align:middle line:84% And in fact, very often, they're not distinct. 00:04:20.170 --> 00:04:21.740 align:middle line:90% So if we choose-- 00:04:21.740 --> 00:04:22.270 align:middle line:90% so here. 00:04:22.270 --> 00:04:24.110 align:middle line:90% 1, 2, 3, 4, 5, 6, 7. 00:04:24.110 --> 00:04:25.250 align:middle line:90% Here are the eight digits. 00:04:25.250 --> 00:04:27.370 align:middle line:84% We're going to choose a number for each. 00:04:27.370 --> 00:04:30.470 align:middle line:84% If I choose a 7 for our first digit, 00:04:30.470 --> 00:04:33.640 align:middle line:90% well, I had 10 options, right? 00:04:33.640 --> 00:04:36.130 align:middle line:84% How many options do I have for the next digit? 00:04:36.130 --> 00:04:38.180 align:middle line:90% 10, right? 00:04:38.180 --> 00:04:40.430 align:middle line:84% I'm allowed to choose 7 again if I want to. 00:04:40.430 --> 00:04:42.500 align:middle line:90% So I have 10 options here. 00:04:42.500 --> 00:04:44.050 align:middle line:90% Maybe I do choose 7 again. 00:04:44.050 --> 00:04:46.150 align:middle line:90% How many options here? 00:04:46.150 --> 00:04:48.280 align:middle line:90% Still 10. 00:04:48.280 --> 00:04:50.150 align:middle line:90% Maybe I choose a 4 this time. 00:04:50.150 --> 00:04:50.840 align:middle line:90% And so on. 00:04:50.840 --> 00:04:53.290 align:middle line:84% I can fill this in every time I have 00:04:53.290 --> 00:04:55.650 align:middle line:90% 10 options for my next digit. 00:04:55.650 --> 00:05:00.110 align:middle line:84% So the number of choices is 10 times 10 times 10, 8 times. 00:05:00.110 --> 00:05:02.540 align:middle line:90% 10 to the eighth. 00:05:02.540 --> 00:05:03.960 align:middle line:90% Does that make sense? 00:05:03.960 --> 00:05:04.520 align:middle line:90% Cool. 00:05:04.520 --> 00:05:06.960 align:middle line:84% This in fact, is just the regular product rule. 00:05:06.960 --> 00:05:09.180 align:middle line:84% We didn't need the generalized product rule here 00:05:09.180 --> 00:05:12.180 align:middle line:84% because the set of choices I have is just all 10 digits. 00:05:12.180 --> 00:05:15.650 align:middle line:84% The set of choices I have is the same, all 10 digits. 00:05:15.650 --> 00:05:18.390 align:middle line:84% The set of choices isn't changing as I make more choices. 00:05:18.390 --> 00:05:20.900 align:middle line:90% Just regular product rule. 00:05:20.900 --> 00:05:24.440 align:middle line:84% I put 8 to the 10 up there just to be a bit of a troll, 00:05:24.440 --> 00:05:29.180 align:middle line:84% because 8 to the 10 and 10 to the 8 are easy to get confused. 00:05:29.180 --> 00:05:31.250 align:middle line:84% And so if you think about it this way, 00:05:31.250 --> 00:05:33.750 align:middle line:84% how many options do I have for this choice, for that choice, 00:05:33.750 --> 00:05:34.740 align:middle line:90% for that choice? 00:05:34.740 --> 00:05:37.160 align:middle line:84% It'll be easy to tell which one of these is right 00:05:37.160 --> 00:05:39.500 align:middle line:90% and which one of them is wrong. 00:05:39.500 --> 00:05:43.250 align:middle line:90% So 10 to the 8 serial numbers. 00:05:43.250 --> 00:05:46.670 align:middle line:84% But we had-- some of these formulas 00:05:46.670 --> 00:05:49.890 align:middle line:84% were sounding like wanted the digits to be distinct. 00:05:49.890 --> 00:05:52.870 align:middle line:84% They're not always, but maybe sometimes they are. 00:05:52.870 --> 00:05:55.350 align:middle line:84% So everyone who pulled out a dollar bill, 00:05:55.350 --> 00:05:57.220 align:middle line:84% can you check your serial number? 00:05:57.220 --> 00:06:00.205 align:middle line:84% And if all of your digits are distinct, 00:06:00.205 --> 00:06:01.205 align:middle line:90% can you raise your hand? 00:06:01.205 --> 00:06:11.940 align:middle line:90% 00:06:11.940 --> 00:06:14.550 align:middle line:90% No one. 00:06:14.550 --> 00:06:16.950 align:middle line:84% Can you raise your hand if you checked a dollar bill 00:06:16.950 --> 00:06:19.170 align:middle line:90% and it failed? 00:06:19.170 --> 00:06:24.660 align:middle line:84% OK, so I'm seeing about 20, 25 people that carry cash. 00:06:24.660 --> 00:06:28.350 align:middle line:84% So out of those 26 bills, none of them 00:06:28.350 --> 00:06:32.300 align:middle line:84% had all distinct digits, which is a shame, 00:06:32.300 --> 00:06:34.300 align:middle line:84% because if someone did have all distinct digits, 00:06:34.300 --> 00:06:35.640 align:middle line:90% I brought candy. 00:06:35.640 --> 00:06:39.600 align:middle line:84% So maybe check the other dollars in your wallet. 00:06:39.600 --> 00:06:42.480 align:middle line:84% But this gives us a natural counting question. 00:06:42.480 --> 00:06:57.900 align:middle line:84% How many serials have all distinct digits? 00:06:57.900 --> 00:06:59.670 align:middle line:84% And let's see if we can count this 00:06:59.670 --> 00:07:04.260 align:middle line:84% using the product rule, or maybe the generalized product rule. 00:07:04.260 --> 00:07:07.335 align:middle line:84% Once again 1, 2, 3, 4, 5, 6, 7, 8. 00:07:07.335 --> 00:07:10.750 align:middle line:84% I like to think about my choices in order one at a time. 00:07:10.750 --> 00:07:13.030 align:middle line:84% So I just draw a little bar to remind myself, 00:07:13.030 --> 00:07:15.240 align:middle line:84% I'm going to choose something here, and remind myself 00:07:15.240 --> 00:07:16.900 align:middle line:90% how many options I had. 00:07:16.900 --> 00:07:20.160 align:middle line:84% So let's see if we can construct serial numbers that 00:07:20.160 --> 00:07:22.620 align:middle line:84% have all distinct digits length eight. 00:07:22.620 --> 00:07:25.140 align:middle line:84% How many options do I have for my first digit? 00:07:25.140 --> 00:07:28.210 align:middle line:90% 10, absolutely. 00:07:28.210 --> 00:07:29.590 align:middle line:90% I have 10 options here. 00:07:29.590 --> 00:07:31.890 align:middle line:90% Maybe this time I chose a 3. 00:07:31.890 --> 00:07:34.590 align:middle line:84% How many options do I have for the second digit? 00:07:34.590 --> 00:07:36.660 align:middle line:90% Nine. 00:07:36.660 --> 00:07:38.490 align:middle line:90% Is it always nine? 00:07:38.490 --> 00:07:41.070 align:middle line:90% What if this digit had been 8? 00:07:41.070 --> 00:07:42.600 align:middle line:90% Still 9, right? 00:07:42.600 --> 00:07:44.850 align:middle line:84% No matter what I did here, there's always nine options 00:07:44.850 --> 00:07:45.790 align:middle line:90% left. 00:07:45.790 --> 00:07:48.330 align:middle line:90% So maybe I choose a 7. 00:07:48.330 --> 00:07:50.600 align:middle line:90% How many options here? 00:07:50.600 --> 00:07:51.110 align:middle line:90% Eight. 00:07:51.110 --> 00:07:51.950 align:middle line:90% I've used two. 00:07:51.950 --> 00:07:53.790 align:middle line:90% I'm not allowed to reuse them. 00:07:53.790 --> 00:07:56.180 align:middle line:90% So there are eight options left. 00:07:56.180 --> 00:08:03.470 align:middle line:84% Then seven, then six, then five, then four, then three. 00:08:03.470 --> 00:08:06.650 align:middle line:84% And therefore, by the generalized product rule-- 00:08:06.650 --> 00:08:15.380 align:middle line:84% by the generalized product rule, the number of length eight 00:08:15.380 --> 00:08:20.750 align:middle line:84% serial numbers with all distinct digits is 10 times 9 times 00:08:20.750 --> 00:08:25.790 align:middle line:84% 8, all the way down to 3, just the product 00:08:25.790 --> 00:08:28.640 align:middle line:84% of the number of choices I have at each step. 00:08:28.640 --> 00:08:31.500 align:middle line:84% Which, by the way, someone had guessed previously. 00:08:31.500 --> 00:08:34.640 align:middle line:90% So well done with that. 00:08:34.640 --> 00:08:37.820 align:middle line:84% And now you could ask, out of all the serials, out of all 10 00:08:37.820 --> 00:08:40.820 align:middle line:84% to the 8 serial numbers, now we know how many of 00:08:40.820 --> 00:08:42.600 align:middle line:90% them have all distinct digits. 00:08:42.600 --> 00:08:44.510 align:middle line:90% What fraction is that? 00:08:44.510 --> 00:08:46.670 align:middle line:90% Well, we just divide. 00:08:46.670 --> 00:08:50.210 align:middle line:84% This number divided by 10 to the eighth 00:08:50.210 --> 00:08:52.970 align:middle line:84% tells us there are actually only about 1.8% 00:08:52.970 --> 00:08:55.550 align:middle line:84% of dollar bills that have all distinct digits. 00:08:55.550 --> 00:08:58.560 align:middle line:90% So about less than 2%. 00:08:58.560 --> 00:09:01.760 align:middle line:84% So since we had like 25 or 30 dollar bills 00:09:01.760 --> 00:09:03.440 align:middle line:84% that we checked collectively, it's 00:09:03.440 --> 00:09:06.950 align:middle line:84% no surprise that we didn't get any of them that 00:09:06.950 --> 00:09:10.010 align:middle line:90% fit this criterion. 00:09:10.010 --> 00:09:11.960 align:middle line:84% If more people carried cash, maybe we'd 00:09:11.960 --> 00:09:14.730 align:middle line:84% be seeing some candy payouts today. 00:09:14.730 --> 00:09:17.540 align:middle line:90% Shame. 00:09:17.540 --> 00:09:21.650 align:middle line:84% Questions about the generalized product rule? 00:09:21.650 --> 00:09:22.400 align:middle line:90% Wonderful. 00:09:22.400 --> 00:09:25.315 align:middle line:84% Let's move on to the next rule, which is the division rule. 00:09:25.315 --> 00:09:31.490 align:middle line:90% 00:09:31.490 --> 00:09:34.130 align:middle line:90% Wonderful. 00:09:34.130 --> 00:09:36.595 align:middle line:84% So, so far, we've seen the bijection rule. 00:09:36.595 --> 00:09:39.650 align:middle line:90% 00:09:39.650 --> 00:09:51.260 align:middle line:84% So if f from a set A to a set B is a bijection, 00:09:51.260 --> 00:09:56.120 align:middle line:84% then the size of A equals the size of B. 00:09:56.120 --> 00:09:58.850 align:middle line:84% A bijection, by the way, is also called a one 00:09:58.850 --> 00:10:00.540 align:middle line:90% to one correspondence. 00:10:00.540 --> 00:10:08.690 align:middle line:84% One to one correspondence, where the picture in your head 00:10:08.690 --> 00:10:10.940 align:middle line:84% should be that every one thing on the left 00:10:10.940 --> 00:10:13.290 align:middle line:84% corresponds to one thing on the right. 00:10:13.290 --> 00:10:15.890 align:middle line:90% They're paired up one to one. 00:10:15.890 --> 00:10:20.370 align:middle line:84% Thinking of this function in our directed arrow drawing fashion, 00:10:20.370 --> 00:10:23.520 align:middle line:84% everything from A has exactly one arrow out, 00:10:23.520 --> 00:10:26.430 align:middle line:84% and everything in B has exactly one arrow in. 00:10:26.430 --> 00:10:28.680 align:middle line:84% That's what it means to be a bijection. 00:10:28.680 --> 00:10:30.570 align:middle line:84% So it's a one to one correspondence. 00:10:30.570 --> 00:10:32.810 align:middle line:90% One arrow out, one arrow in. 00:10:32.810 --> 00:10:35.780 align:middle line:84% We can generalize this a little bit. 00:10:35.780 --> 00:10:56.370 align:middle line:84% So if instead every a in A has exactly one arrow out 00:10:56.370 --> 00:11:09.150 align:middle line:84% and every b in B has exactly k arrows in, 00:11:09.150 --> 00:11:14.550 align:middle line:84% then we can conclude that the size of A divided by k 00:11:14.550 --> 00:11:20.083 align:middle line:84% equals the size of B. So here's my picture. 00:11:20.083 --> 00:11:27.660 align:middle line:90% 00:11:27.660 --> 00:11:30.460 align:middle line:84% Everything on the left has one arrow out. 00:11:30.460 --> 00:11:34.020 align:middle line:84% Everything on the right has two arrows in, exactly two arrows 00:11:34.020 --> 00:11:35.880 align:middle line:90% in every time. 00:11:35.880 --> 00:11:38.040 align:middle line:84% Then there are double as many elements on the left 00:11:38.040 --> 00:11:40.650 align:middle line:90% as there are on the right. 00:11:40.650 --> 00:11:45.900 align:middle line:84% If it were three arrows in, if everything on the right 00:11:45.900 --> 00:11:47.350 align:middle line:84% has exactly three arrows in, then 00:11:47.350 --> 00:11:49.350 align:middle line:84% there are three times as many things on the left 00:11:49.350 --> 00:11:51.540 align:middle line:90% as there are on the right. 00:11:51.540 --> 00:11:54.340 align:middle line:84% So this is a generalization of a bijection, 00:11:54.340 --> 00:11:56.760 align:middle line:84% or a one to one correspondence to what's 00:11:56.760 --> 00:11:58.796 align:middle line:90% called a k to 1 correspondence. 00:11:58.796 --> 00:12:04.740 align:middle line:90% 00:12:04.740 --> 00:12:07.440 align:middle line:84% Because what this ends up doing is saying that we've divided 00:12:07.440 --> 00:12:09.870 align:middle line:84% the left side into groups of size k 00:12:09.870 --> 00:12:13.320 align:middle line:84% that correspond to each one thing on the right. 00:12:13.320 --> 00:12:19.380 align:middle line:84% So this one thing on the right matches 1, 2, 3 things 00:12:19.380 --> 00:12:20.500 align:middle line:90% on the left. 00:12:20.500 --> 00:12:22.710 align:middle line:84% And everything on the right is paired 00:12:22.710 --> 00:12:26.160 align:middle line:84% with three distinct things on the left. 00:12:26.160 --> 00:12:29.130 align:middle line:84% And so this is a useful tool for counting that generalizes 00:12:29.130 --> 00:12:30.190 align:middle line:90% the bijection rule. 00:12:30.190 --> 00:12:33.900 align:middle line:84% We don't have to match things up one to one as long as we know 00:12:33.900 --> 00:12:36.850 align:middle line:84% that we're matching the same number of things every time, 00:12:36.850 --> 00:12:39.660 align:middle line:84% and then we just divide by that thing. 00:12:39.660 --> 00:12:42.200 align:middle line:84% All right, let's put this to use. 00:12:42.200 --> 00:12:45.610 align:middle line:90% 00:12:45.610 --> 00:12:46.813 align:middle line:90% So example. 00:12:46.813 --> 00:12:49.870 align:middle line:90% 00:12:49.870 --> 00:12:51.225 align:middle line:90% Knights of the Round Table. 00:12:51.225 --> 00:12:58.510 align:middle line:90% 00:12:58.510 --> 00:13:02.123 align:middle line:84% So the setup here is we're going to have n knights. 00:13:02.123 --> 00:13:05.440 align:middle line:90% 00:13:05.440 --> 00:13:07.750 align:middle line:84% Maybe it should have been k knights. n knights 00:13:07.750 --> 00:13:19.330 align:middle line:84% all sitting around table around a round table at n equally 00:13:19.330 --> 00:13:20.310 align:middle line:90% spaced seats. 00:13:20.310 --> 00:13:27.940 align:middle line:90% 00:13:27.940 --> 00:13:32.590 align:middle line:84% How many different arrangements are 00:13:32.590 --> 00:13:54.050 align:middle line:84% possible if rotations of a seating assignment 00:13:54.050 --> 00:13:58.266 align:middle line:90% are considered equivalent? 00:13:58.266 --> 00:14:01.490 align:middle line:90% 00:14:01.490 --> 00:14:04.530 align:middle line:84% So n knights sitting around a table. 00:14:04.530 --> 00:14:06.282 align:middle line:84% If you have one seating assignment 00:14:06.282 --> 00:14:08.240 align:middle line:84% and you have a second seating assignment that's 00:14:08.240 --> 00:14:10.550 align:middle line:84% just a rotation of the first, then we 00:14:10.550 --> 00:14:12.650 align:middle line:84% want to consider those two assignments the same 00:14:12.650 --> 00:14:13.980 align:middle line:90% seating assignment. 00:14:13.980 --> 00:14:16.380 align:middle line:84% We don't care which exact seats they're in, 00:14:16.380 --> 00:14:20.220 align:middle line:84% as long as the cyclic order around the circle is the same. 00:14:20.220 --> 00:14:25.770 align:middle line:84% So for example, if n is 4 and we have 1 here, 00:14:25.770 --> 00:14:28.680 align:middle line:84% then 3 here, then 4 here and 2 here, 00:14:28.680 --> 00:14:32.030 align:middle line:84% that's one possible seating assignment. 00:14:32.030 --> 00:14:33.660 align:middle line:90% Well, here's a different one. 00:14:33.660 --> 00:14:37.190 align:middle line:90% 1 here, 3 here, 4 here, 2 here. 00:14:37.190 --> 00:14:39.728 align:middle line:84% But this assignment is the same as that assignment 00:14:39.728 --> 00:14:41.270 align:middle line:84% because we've just rotated the table. 00:14:41.270 --> 00:14:43.980 align:middle line:84% And we're just saying we don't care about rotating the table. 00:14:43.980 --> 00:14:47.300 align:middle line:84% So in fact, we want to consider these to be equivalent seating 00:14:47.300 --> 00:14:48.980 align:middle line:90% assignments. 00:14:48.980 --> 00:14:53.010 align:middle line:84% Just like we want to consider this to be equivalent to 1, 00:14:53.010 --> 00:14:57.690 align:middle line:84% then 3, then 4, then 2 this way, and equivalent to 1, 00:14:57.690 --> 00:15:00.390 align:middle line:90% then 3, then 4, then 2 this way. 00:15:00.390 --> 00:15:03.590 align:middle line:84% So all four of these rotated seating assignments 00:15:03.590 --> 00:15:06.030 align:middle line:84% are equivalent because it's the same order, 00:15:06.030 --> 00:15:08.660 align:middle line:84% they're just starting in a different place. 00:15:08.660 --> 00:15:10.610 align:middle line:90% Does that make sense? 00:15:10.610 --> 00:15:11.120 align:middle line:90% Cool. 00:15:11.120 --> 00:15:12.390 align:middle line:90% So that's our question. 00:15:12.390 --> 00:15:15.290 align:middle line:84% How many different seating arrangements for these n knights 00:15:15.290 --> 00:15:20.990 align:middle line:84% are possible, given that we don't care about rotations? 00:15:20.990 --> 00:15:24.930 align:middle line:84% And to solve it, let's see if we can use a k to 1 map, 00:15:24.930 --> 00:15:26.750 align:middle line:84% because that's what we're trying to learn 00:15:26.750 --> 00:15:28.195 align:middle line:90% at this point in this lecture. 00:15:28.195 --> 00:15:31.010 align:middle line:90% 00:15:31.010 --> 00:15:34.940 align:middle line:84% So here is the map I'm going to recommend. 00:15:34.940 --> 00:15:46.710 align:middle line:84% Let's let A be the set of permutations of 1 through n. 00:15:46.710 --> 00:15:48.910 align:middle line:84% Permutation just means a reordering. 00:15:48.910 --> 00:15:52.890 align:middle line:84% So it's all of the orders of the set 1 through n. 00:15:52.890 --> 00:16:05.850 align:middle line:84% And B is the set of cyclic seating arrangements. 00:16:05.850 --> 00:16:08.850 align:middle line:84% So B is the thing we're trying to count, 00:16:08.850 --> 00:16:12.030 align:middle line:84% how many different seating arrangements are possible. 00:16:12.030 --> 00:16:14.260 align:middle line:90% Well A, we know how to count. 00:16:14.260 --> 00:16:15.560 align:middle line:90% What's the size of A? 00:16:15.560 --> 00:16:21.930 align:middle line:90% 00:16:21.930 --> 00:16:22.753 align:middle line:90% What is it? 00:16:22.753 --> 00:16:23.670 align:middle line:90% AUDIENCE: n factorial. 00:16:23.670 --> 00:16:25.440 align:middle line:90% ZACHARY ABEL: n factorial. 00:16:25.440 --> 00:16:27.658 align:middle line:90% That's exactly right. 00:16:27.658 --> 00:16:29.700 align:middle line:84% That was our first example of generalized product 00:16:29.700 --> 00:16:30.930 align:middle line:90% rule on Tuesday. 00:16:30.930 --> 00:16:33.580 align:middle line:84% There are 52 factorial orders for a deck of cards. 00:16:33.580 --> 00:16:37.720 align:middle line:84% There are n factorial orders for a set of n knights. 00:16:37.720 --> 00:16:38.750 align:middle line:90% All right. 00:16:38.750 --> 00:16:41.420 align:middle line:84% So if we want to use the division rule, 00:16:41.420 --> 00:16:44.650 align:middle line:84% we have a set, A, where we know the size. 00:16:44.650 --> 00:16:48.400 align:middle line:84% And we need to be able to define a function from A to the set B 00:16:48.400 --> 00:16:50.830 align:middle line:84% that we care about and make sure that it's 00:16:50.830 --> 00:16:53.800 align:middle line:90% a k to 1 map for some k. 00:16:53.800 --> 00:16:55.580 align:middle line:84% And here's the map we're going to do. 00:16:55.580 --> 00:17:00.370 align:middle line:84% So f is going to take in a permutation, so an element of a. 00:17:00.370 --> 00:17:03.614 align:middle line:90% So a1, a2 up to a n. 00:17:03.614 --> 00:17:06.220 align:middle line:90% 00:17:06.220 --> 00:17:09.579 align:middle line:84% So it takes in a permutation, an element of A, 00:17:09.579 --> 00:17:12.230 align:middle line:84% and it needs to spit out an element of B. 00:17:12.230 --> 00:17:15.859 align:middle line:84% So it needs to spit out a cyclic ordering of these numbers. 00:17:15.859 --> 00:17:21.339 align:middle line:84% And it's going to spit out this cyclic ordering that 00:17:21.339 --> 00:17:25.030 align:middle line:84% just sets them in that order around the table. 00:17:25.030 --> 00:17:31.480 align:middle line:84% Yeah, the most straightforward thing I can think of to do here. 00:17:31.480 --> 00:17:34.300 align:middle line:84% Now that we have this function, let's 00:17:34.300 --> 00:17:35.775 align:middle line:90% check if it's a k to 1 map. 00:17:35.775 --> 00:17:41.470 align:middle line:90% 00:17:41.470 --> 00:17:44.770 align:middle line:84% So what does it mean for f to be a k to 1 map? 00:17:44.770 --> 00:17:50.650 align:middle line:90% Is f a k to 1 map? 00:17:50.650 --> 00:17:55.120 align:middle line:84% Well, it means that everything in B 00:17:55.120 --> 00:18:00.310 align:middle line:84% needs to come from exactly k elements of A. 00:18:00.310 --> 00:18:16.540 align:middle line:84% So we want every cyclic order comes from exactly k 00:18:16.540 --> 00:18:19.180 align:middle line:90% permutations. 00:18:19.180 --> 00:18:24.400 align:middle line:84% And when we say "comes from," we mean via this function f. 00:18:24.400 --> 00:18:27.910 align:middle line:90% Via function f. 00:18:27.910 --> 00:18:31.210 align:middle line:84% So as an example, maybe n is five this time 00:18:31.210 --> 00:18:37.670 align:middle line:84% and we have the order 4, 5, 2, 3, 1, some cyclic order here. 00:18:37.670 --> 00:18:41.310 align:middle line:84% How many different permutations of 1 through 5 00:18:41.310 --> 00:18:43.348 align:middle line:90% map to this cyclic ordering? 00:18:43.348 --> 00:18:44.515 align:middle line:90% Who can give me one of them? 00:18:44.515 --> 00:18:54.230 align:middle line:90% 00:18:54.230 --> 00:18:57.500 align:middle line:84% So we have this function f that takes a permutation of 1 00:18:57.500 --> 00:19:00.260 align:middle line:84% through n and just lays it out around the circle. 00:19:00.260 --> 00:19:03.770 align:middle line:84% So who can give me a permutation that becomes this once you 00:19:03.770 --> 00:19:06.680 align:middle line:90% lay it out around the circle? 00:19:06.680 --> 00:19:08.790 align:middle line:84% Let's see if we can get some new hands. 00:19:08.790 --> 00:19:10.080 align:middle line:90% Yes, please. 00:19:10.080 --> 00:19:11.080 align:middle line:90% AUDIENCE: 1, 4, 5, 2, 3. 00:19:11.080 --> 00:19:13.580 align:middle line:90% ZACHARY ABEL: 1, 4, 5, 2, 3. 00:19:13.580 --> 00:19:16.520 align:middle line:90% 1, 4, 5, 2, 3. 00:19:16.520 --> 00:19:18.220 align:middle line:84% If we lay that around a circle, we're 00:19:18.220 --> 00:19:20.040 align:middle line:90% going to get 1, 4, 5, 2, 3. 00:19:20.040 --> 00:19:20.880 align:middle line:90% Looks good to me. 00:19:20.880 --> 00:19:21.380 align:middle line:90% Excellent. 00:19:21.380 --> 00:19:23.240 align:middle line:90% Who can give me another? 00:19:23.240 --> 00:19:24.440 align:middle line:90% Yes? 00:19:24.440 --> 00:19:25.940 align:middle line:90% AUDIENCE: 4, 5 2, 3 1. 00:19:25.940 --> 00:19:27.440 align:middle line:90% ZACHARY ABEL: 4, 5, 2, 3, 1. 00:19:27.440 --> 00:19:31.430 align:middle line:84% 4, 5, 2, 3, 1 If we lay that around a circle, 00:19:31.430 --> 00:19:34.380 align:middle line:84% we're going to get 4, 5, 2, 3, 1 Looks good to me. 00:19:34.380 --> 00:19:36.230 align:middle line:90% Same cyclic ordering. 00:19:36.230 --> 00:19:39.120 align:middle line:84% All we're doing is choosing a starting point. 00:19:39.120 --> 00:19:41.997 align:middle line:84% If we know what cyclic ordering we're aiming for, 00:19:41.997 --> 00:19:43.580 align:middle line:84% well, we just have to choose who we're 00:19:43.580 --> 00:19:47.420 align:middle line:84% going to start with and then list the rest of them 00:19:47.420 --> 00:19:49.230 align:middle line:90% in order around the circle. 00:19:49.230 --> 00:19:50.810 align:middle line:84% And that's how we can reconstruct 00:19:50.810 --> 00:19:53.490 align:middle line:84% all of the permutations that map to the cyclic order. 00:19:53.490 --> 00:19:54.980 align:middle line:84% If we start with 2, then it's going 00:19:54.980 --> 00:19:58.760 align:middle line:90% to look like 2, 3, 1, 4, 5. 00:19:58.760 --> 00:20:00.920 align:middle line:84% If, instead, we start with 3, it's 00:20:00.920 --> 00:20:05.570 align:middle line:84% going to look like 3, 1, 4, 5, 2. 00:20:05.570 --> 00:20:09.710 align:middle line:84% And the only one we haven't done yet is starting with 5. 00:20:09.710 --> 00:20:13.682 align:middle line:90% 5, 2, 3, 1, 4. 00:20:13.682 --> 00:20:15.570 align:middle line:90% 5, 2, 3, 1, 4. 00:20:15.570 --> 00:20:18.560 align:middle line:90% And I claim that's all of them. 00:20:18.560 --> 00:20:21.230 align:middle line:84% So in this case, for this particular cyclic ordering, 00:20:21.230 --> 00:20:24.040 align:middle line:84% there were five permutations that mapped to it. 00:20:24.040 --> 00:20:26.960 align:middle line:90% 00:20:26.960 --> 00:20:37.950 align:middle line:84% In general, I claim for each cyclic ordering, 00:20:37.950 --> 00:20:44.580 align:middle line:84% there exists exactly n permutations that 00:20:44.580 --> 00:20:54.070 align:middle line:84% map to it for the same reason, we just saw for this example. 00:20:54.070 --> 00:20:55.590 align:middle line:84% The only additional information we 00:20:55.590 --> 00:20:58.620 align:middle line:84% need to reconstruct all of these permutations 00:20:58.620 --> 00:21:05.670 align:middle line:84% is which starting point we're choosing, because just 00:21:05.670 --> 00:21:13.790 align:middle line:90% need to pick a starting point. 00:21:13.790 --> 00:21:17.280 align:middle line:90% 00:21:17.280 --> 00:21:20.370 align:middle line:84% So for every cyclic ordering, there 00:21:20.370 --> 00:21:22.860 align:middle line:84% exists exactly n permutations that 00:21:22.860 --> 00:21:26.130 align:middle line:84% map to it, which is exactly the condition we 00:21:26.130 --> 00:21:28.965 align:middle line:90% need for f to be an n to 1 map. 00:21:28.965 --> 00:21:32.681 align:middle line:90% 00:21:32.681 --> 00:21:36.230 align:middle line:90% n to 1 map. 00:21:36.230 --> 00:21:39.400 align:middle line:84% Everything on the right comes from exactly n things 00:21:39.400 --> 00:21:41.230 align:middle line:90% on the left. 00:21:41.230 --> 00:21:44.355 align:middle line:84% And what do we conclude now that we know f is an n to 1 map? 00:21:44.355 --> 00:21:53.560 align:middle line:90% 00:21:53.560 --> 00:21:57.000 align:middle line:84% Why are k to one maps in general helpful? 00:21:57.000 --> 00:22:00.520 align:middle line:90% 00:22:00.520 --> 00:22:01.360 align:middle line:90% Yes? 00:22:01.360 --> 00:22:02.860 align:middle line:84% AUDIENCE: [INAUDIBLE] the size of B. 00:22:02.860 --> 00:22:04.402 align:middle line:84% ZACHARY ABEL: Yeah, we can understand 00:22:04.402 --> 00:22:06.760 align:middle line:84% the size of B. k to 1 maps are helpful because they 00:22:06.760 --> 00:22:10.630 align:middle line:84% let us relate the size of A to the size of B. 00:22:10.630 --> 00:22:13.930 align:middle line:84% And in this example, we know the size of A. 00:22:13.930 --> 00:22:15.830 align:middle line:90% We know it's an n to 1 map. 00:22:15.830 --> 00:22:23.050 align:middle line:84% And therefore, we know the size of B. 00:22:23.050 --> 00:22:25.810 align:middle line:84% So the size of B is just the size 00:22:25.810 --> 00:22:31.330 align:middle line:84% of A divided by n, because we decided it's an n to 1 map. 00:22:31.330 --> 00:22:34.750 align:middle line:84% And therefore, the size of B is n factorial 00:22:34.750 --> 00:22:40.570 align:middle line:84% divided by n, also known as n minus 1 factorial. 00:22:40.570 --> 00:22:44.120 align:middle line:90% Does that make sense? 00:22:44.120 --> 00:22:44.620 align:middle line:90% Cool. 00:22:44.620 --> 00:22:46.630 align:middle line:90% I see some nods. 00:22:46.630 --> 00:22:48.260 align:middle line:90% I see some blank stares again. 00:22:48.260 --> 00:22:49.490 align:middle line:90% That's fine. 00:22:49.490 --> 00:22:52.300 align:middle line:84% Ask questions when you have them. 00:22:52.300 --> 00:22:59.080 align:middle line:84% OK, let's look at one more example of the division rule. 00:22:59.080 --> 00:23:01.610 align:middle line:84% I want to look at increasing serial numbers again, 00:23:01.610 --> 00:23:02.990 align:middle line:90% but this time just length three. 00:23:02.990 --> 00:23:04.960 align:middle line:84% Let's make it easier on ourselves. 00:23:04.960 --> 00:23:16.540 align:middle line:84% So the number of serial numbers length three. 00:23:16.540 --> 00:23:19.480 align:middle line:84% And now I want to know how many of them have 00:23:19.480 --> 00:23:21.315 align:middle line:90% all three digits that are in-- 00:23:21.315 --> 00:23:23.860 align:middle line:90% 00:23:23.860 --> 00:23:25.020 align:middle line:90% sorry, one moment. 00:23:25.020 --> 00:23:28.670 align:middle line:90% 00:23:28.670 --> 00:23:30.240 align:middle line:90% Sorry, I got ahead of myself. 00:23:30.240 --> 00:23:31.020 align:middle line:90% Different example. 00:23:31.020 --> 00:23:32.930 align:middle line:84% Still length three serial numbers, 00:23:32.930 --> 00:23:34.310 align:middle line:90% I just want to count these. 00:23:34.310 --> 00:23:35.310 align:middle line:90% That's all I want to do. 00:23:35.310 --> 00:23:37.340 align:middle line:84% How many length three serial numbers 00:23:37.340 --> 00:23:40.473 align:middle line:90% are there with distinct digits? 00:23:40.473 --> 00:23:52.130 align:middle line:90% 00:23:52.130 --> 00:23:55.170 align:middle line:84% And we already have one way that we know to count this. 00:23:55.170 --> 00:23:56.990 align:middle line:84% We have the generalized product rule method 00:23:56.990 --> 00:23:58.240 align:middle line:90% that we used for length eight. 00:23:58.240 --> 00:23:59.910 align:middle line:84% We can do the same for length three. 00:23:59.910 --> 00:24:02.000 align:middle line:84% I want to propose a different method that 00:24:02.000 --> 00:24:04.460 align:middle line:90% uses the division rule instead. 00:24:04.460 --> 00:24:07.670 align:middle line:90% And here's what I'm going to do. 00:24:07.670 --> 00:24:16.010 align:middle line:84% Let's let P be the set of permutations of 1 through 10-- 00:24:16.010 --> 00:24:17.230 align:middle line:90% sorry, of 0 through 9. 00:24:17.230 --> 00:24:20.840 align:middle line:90% 00:24:20.840 --> 00:24:29.040 align:middle line:84% And let's let S be the set of length three serials 00:24:29.040 --> 00:24:32.540 align:middle line:90% with distinct digits. 00:24:32.540 --> 00:24:37.560 align:middle line:90% 00:24:37.560 --> 00:24:38.210 align:middle line:90% All right. 00:24:38.210 --> 00:24:45.540 align:middle line:90% 00:24:45.540 --> 00:24:47.050 align:middle line:84% And I just checked my notes again. 00:24:47.050 --> 00:24:49.800 align:middle line:90% I got myself confused yet again. 00:24:49.800 --> 00:24:50.770 align:middle line:90% Forgive me. 00:24:50.770 --> 00:24:56.460 align:middle line:84% Let me change the details of the example just a little bit. 00:24:56.460 --> 00:24:59.520 align:middle line:84% I want the number of serials of length three with not only 00:24:59.520 --> 00:25:03.230 align:middle line:84% distinct digits, but actually digits in increasing order. 00:25:03.230 --> 00:25:10.770 align:middle line:90% 00:25:10.770 --> 00:25:15.390 align:middle line:90% So I'm OK with 2 and 4 and 9. 00:25:15.390 --> 00:25:19.308 align:middle line:90% I'm OK with 0, 1, 3. 00:25:19.308 --> 00:25:20.910 align:middle line:90% 0, 1, 2 is fine. 00:25:20.910 --> 00:25:26.940 align:middle line:84% I'm not OK with 4, 2, 3, because 4 and 2 are in the wrong order. 00:25:26.940 --> 00:25:32.430 align:middle line:84% I'm not OK with 2, 2, 5 because I want strictly increasing. 00:25:32.430 --> 00:25:38.790 align:middle line:84% So how many length three serials have strictly increasing digits? 00:25:38.790 --> 00:25:41.100 align:middle line:90% Is the question clear? 00:25:41.100 --> 00:25:44.790 align:middle line:90% OK, thank you. 00:25:44.790 --> 00:25:47.020 align:middle line:84% And let's do the same thing we were doing before. 00:25:47.020 --> 00:25:50.260 align:middle line:84% Let P be the set of permutations of 0 through 9 00:25:50.260 --> 00:25:53.070 align:middle line:84% and let S be the set of length three serials 00:25:53.070 --> 00:25:57.120 align:middle line:90% with increasing digits. 00:25:57.120 --> 00:25:59.220 align:middle line:90% Strictly increasing digits. 00:25:59.220 --> 00:26:01.800 align:middle line:84% And let's see if we can relate P with S in order 00:26:01.800 --> 00:26:04.977 align:middle line:90% to count the size of S. 00:26:04.977 --> 00:26:06.310 align:middle line:90% And here's what I'm going to do. 00:26:06.310 --> 00:26:07.860 align:middle line:90% Let's define a function. 00:26:07.860 --> 00:26:13.740 align:middle line:84% Let's define a function, g, from P to S. 00:26:13.740 --> 00:26:16.110 align:middle line:84% And it's going to look like this. 00:26:16.110 --> 00:26:23.080 align:middle line:84% g of a permutation a0, a1, a2 up to a9. 00:26:23.080 --> 00:26:26.530 align:middle line:90% That's a generic element of P. 00:26:26.530 --> 00:26:33.043 align:middle line:84% That's going to go to the serial that has a0, a1, a2 sorted. 00:26:33.043 --> 00:26:36.310 align:middle line:90% 00:26:36.310 --> 00:26:38.280 align:middle line:84% So take those three digits and them. 00:26:38.280 --> 00:26:41.060 align:middle line:90% 00:26:41.060 --> 00:26:44.830 align:middle line:84% That's a way to turn a permutation into a serial number 00:26:44.830 --> 00:26:45.870 align:middle line:90% with increasing digits. 00:26:45.870 --> 00:26:48.520 align:middle line:90% 00:26:48.520 --> 00:26:51.790 align:middle line:84% By the way, a different way to think about this problem. 00:26:51.790 --> 00:26:54.130 align:middle line:90% You can think of S-- 00:26:54.130 --> 00:27:06.880 align:middle line:84% S is equivalently the set of size three subsets of 0 00:27:06.880 --> 00:27:08.260 align:middle line:90% through 9. 00:27:08.260 --> 00:27:11.300 align:middle line:84% If we know they have to come in increasing order, 00:27:11.300 --> 00:27:13.433 align:middle line:84% then all we care about is which three we chose. 00:27:13.433 --> 00:27:17.920 align:middle line:90% 00:27:17.920 --> 00:27:20.440 align:middle line:84% So if I tell you my serial has increasing order 00:27:20.440 --> 00:27:24.220 align:middle line:84% and has the digits of 5 and 4 and 9, 00:27:24.220 --> 00:27:27.160 align:middle line:84% then I know it has to be 4, 5, 9. 00:27:27.160 --> 00:27:29.380 align:middle line:84% So just the set of three digits that we chose 00:27:29.380 --> 00:27:31.990 align:middle line:84% is enough to know which serial it is. 00:27:31.990 --> 00:27:35.080 align:middle line:84% So if we want to think about this as instead the set of size 00:27:35.080 --> 00:27:40.750 align:middle line:84% three subsets, then this function g would look like this, 00:27:40.750 --> 00:27:43.955 align:middle line:84% would just go to the set a0, a1, a2. 00:27:43.955 --> 00:27:47.890 align:middle line:90% 00:27:47.890 --> 00:27:51.040 align:middle line:84% So this is a way to map permutations of the digits 0 00:27:51.040 --> 00:27:53.490 align:middle line:84% through 9 to subsets of size three. 00:27:53.490 --> 00:27:56.650 align:middle line:90% 00:27:56.650 --> 00:27:59.360 align:middle line:84% And now let's see if this g is a helpful function. 00:27:59.360 --> 00:28:02.890 align:middle line:84% Let's see if it's k to 1 for some k. 00:28:02.890 --> 00:28:06.670 align:middle line:84% So the question becomes, for every size three 00:28:06.670 --> 00:28:12.340 align:middle line:84% subset of the digits, how many permutations map to it? 00:28:12.340 --> 00:28:23.552 align:middle line:84% For example, how many permutations map to 3, 6, 7? 00:28:23.552 --> 00:28:26.690 align:middle line:90% 00:28:26.690 --> 00:28:29.720 align:middle line:84% One permutation maps to this set. 00:28:29.720 --> 00:28:32.123 align:middle line:90% That was one suggestion. 00:28:32.123 --> 00:28:33.540 align:middle line:84% Anyone have different suggestions? 00:28:33.540 --> 00:28:34.373 align:middle line:90% Is it more than one? 00:28:34.373 --> 00:28:37.370 align:middle line:90% 00:28:37.370 --> 00:28:38.913 align:middle line:90% In the back. 00:28:38.913 --> 00:28:39.830 align:middle line:90% AUDIENCE: 3 factorial. 00:28:39.830 --> 00:28:42.920 align:middle line:90% ZACHARY ABEL: 3 factorial, OK. 00:28:42.920 --> 00:28:46.350 align:middle line:84% So we have 1, we have 3 factorial. 00:28:46.350 --> 00:28:47.970 align:middle line:90% Any other suggestions? 00:28:47.970 --> 00:28:49.100 align:middle line:90% Yes. 00:28:49.100 --> 00:28:50.810 align:middle line:84% AUDIENCE: 7 factorial times 3 factorial. 00:28:50.810 --> 00:28:53.366 align:middle line:84% ZACHARY ABEL: 7 factorial times 3 factorial. 00:28:53.366 --> 00:28:55.490 align:middle line:90% 00:28:55.490 --> 00:28:57.740 align:middle line:84% I'm going to put a question mark next to each of these 00:28:57.740 --> 00:28:59.280 align:middle line:90% because there are guesses. 00:28:59.280 --> 00:29:01.490 align:middle line:84% Let's see if we can figure out which one is right. 00:29:01.490 --> 00:29:06.080 align:middle line:84% So let's see how many permutations map us to this set. 00:29:06.080 --> 00:29:09.882 align:middle line:84% Well, we need to pick some permutation, 1, 2, 3, 4, 5, 6, 00:29:09.882 --> 00:29:13.920 align:middle line:84% 7, 8, 10, some permutation of the 10 digits. 00:29:13.920 --> 00:29:15.883 align:middle line:84% What do we know about these three digits? 00:29:15.883 --> 00:29:19.440 align:middle line:90% 00:29:19.440 --> 00:29:23.520 align:middle line:84% Our function says, you take the first three entries 00:29:23.520 --> 00:29:25.440 align:middle line:90% and you form them into a set. 00:29:25.440 --> 00:29:27.870 align:middle line:90% And that's the output. 00:29:27.870 --> 00:29:31.360 align:middle line:84% So if our output is going to be 3, 6, 7, 00:29:31.360 --> 00:29:33.000 align:middle line:84% then these first three entries have 00:29:33.000 --> 00:29:36.810 align:middle line:90% to be 3, 6, and 7 in some order. 00:29:36.810 --> 00:29:38.070 align:middle line:90% Yeah? 00:29:38.070 --> 00:29:43.020 align:middle line:90% 3, 6, 7 in some order. 00:29:43.020 --> 00:29:45.545 align:middle line:84% I don't care which order, because I'm just 00:29:45.545 --> 00:29:47.920 align:middle line:84% going to take the three of them and plop them into a set. 00:29:47.920 --> 00:29:51.370 align:middle line:84% And a set doesn't care what order they are. 00:29:51.370 --> 00:29:54.360 align:middle line:84% So I know that 3, 6, and 7 have to be in 00:29:54.360 --> 00:29:57.640 align:middle line:84% have to be in some order at the beginning of my permutation. 00:29:57.640 --> 00:29:59.290 align:middle line:84% What about the end of the permutation? 00:29:59.290 --> 00:30:02.160 align:middle line:90% What do I know about this? 00:30:02.160 --> 00:30:04.210 align:middle line:84% What constraints do I have on it, 00:30:04.210 --> 00:30:07.390 align:middle line:84% knowing that G of this permutation equals this set? 00:30:07.390 --> 00:30:07.930 align:middle line:90% Yeah? 00:30:07.930 --> 00:30:09.660 align:middle line:84% AUDIENCE: You can't have 3, 6, or 7. 00:30:09.660 --> 00:30:10.493 align:middle line:90% ZACHARY ABEL: Right. 00:30:10.493 --> 00:30:12.240 align:middle line:90% You can't have 3, 6, or 7. 00:30:12.240 --> 00:30:15.190 align:middle line:84% You have to have all the other digits in some order. 00:30:15.190 --> 00:30:18.870 align:middle line:90% But anything goes, yeah. 00:30:18.870 --> 00:30:28.080 align:middle line:84% So it's the other seven digits in some order. 00:30:28.080 --> 00:30:31.860 align:middle line:84% This is exactly the set of permutations 00:30:31.860 --> 00:30:35.080 align:middle line:84% where g of that permutation gives the set 3, 6, 7. 00:30:35.080 --> 00:30:37.243 align:middle line:84% So we have to put 3, 6, and 7 in some order 00:30:37.243 --> 00:30:39.660 align:middle line:84% at the beginning, put the other seven digits in some order 00:30:39.660 --> 00:30:40.540 align:middle line:90% at the end. 00:30:40.540 --> 00:30:42.820 align:middle line:84% And how many ways are there to do that? 00:30:42.820 --> 00:30:44.560 align:middle line:90% Well, this is product rule. 00:30:44.560 --> 00:30:47.670 align:middle line:84% It's 3 factorial here on the left, 7 factorial here 00:30:47.670 --> 00:30:48.870 align:middle line:90% on the right. 00:30:48.870 --> 00:30:54.070 align:middle line:84% So 3 factorial times 7 factorial is indeed 00:30:54.070 --> 00:30:58.740 align:middle line:84% the number of permutations that map to this set. 00:30:58.740 --> 00:31:00.670 align:middle line:90% Was that's specific to this set? 00:31:00.670 --> 00:31:02.710 align:middle line:84% If I chose a different set of three digits, 00:31:02.710 --> 00:31:04.906 align:middle line:84% would I get a different number of pre-images? 00:31:04.906 --> 00:31:09.450 align:middle line:90% 00:31:09.450 --> 00:31:14.910 align:middle line:84% Like, if I instead chose 0 and 5 and 8? 00:31:14.910 --> 00:31:17.120 align:middle line:84% Well, it would be the same constraint, right? 00:31:17.120 --> 00:31:19.790 align:middle line:84% I would have to have 0, 5 and 8 in some order here, 00:31:19.790 --> 00:31:22.310 align:middle line:84% and then the other seven digits in some order here. 00:31:22.310 --> 00:31:26.830 align:middle line:84% It's still going to be 3 factorial times 7 factorial. 00:31:26.830 --> 00:31:36.340 align:middle line:84% So in fact, g is k to 1, where k equals 00:31:36.340 --> 00:31:41.500 align:middle line:90% 3 factorial times 7 factorial. 00:31:41.500 --> 00:31:42.700 align:middle line:90% Yeah? 00:31:42.700 --> 00:31:45.850 align:middle line:84% Because the same argument we did for this particular example, 00:31:45.850 --> 00:31:47.280 align:middle line:84% in fact, works for every example. 00:31:47.280 --> 00:31:49.930 align:middle line:90% 00:31:49.930 --> 00:31:50.440 align:middle line:90% Great. 00:31:50.440 --> 00:31:52.400 align:middle line:90% And since k is-- 00:31:52.400 --> 00:31:54.830 align:middle line:84% since this map is k to 1 and we know k, 00:31:54.830 --> 00:31:58.780 align:middle line:84% we can compute the size of our various sets. 00:31:58.780 --> 00:32:01.910 align:middle line:84% On one hand, we have permutations of 0 through 9. 00:32:01.910 --> 00:32:04.900 align:middle line:90% So that was 10 factorial. 00:32:04.900 --> 00:32:08.840 align:middle line:84% We put it through a k to 1 map where k is this. 00:32:08.840 --> 00:32:13.450 align:middle line:84% So 3 factorial times 7 factorial in the denominator. 00:32:13.450 --> 00:32:23.005 align:middle line:84% And this equals the size of the set S. Sound good? 00:32:23.005 --> 00:32:26.540 align:middle line:90% 00:32:26.540 --> 00:32:30.890 align:middle line:84% So we mapped from permutations to the set of the first three 00:32:30.890 --> 00:32:31.465 align:middle line:90% elements. 00:32:31.465 --> 00:32:32.840 align:middle line:84% And we decided that every time we 00:32:32.840 --> 00:32:39.650 align:middle line:84% do that, we lose by a factor of 3 factorial times 7 factorial. 00:32:39.650 --> 00:32:45.590 align:middle line:84% And so this is the number of size 3 subsets of 10 objects. 00:32:45.590 --> 00:32:50.190 align:middle line:84% This, by the way, is an extremely useful formula. 00:32:50.190 --> 00:32:52.020 align:middle line:90% Let me generalize this. 00:32:52.020 --> 00:33:10.160 align:middle line:84% So the number of subsets of size k from a set of size n is-- 00:33:10.160 --> 00:33:12.890 align:middle line:84% well, we can do the same kind of argument here. 00:33:12.890 --> 00:33:16.170 align:middle line:84% Map permutations, n factorial of them, 00:33:16.170 --> 00:33:18.990 align:middle line:84% where we take the first k elements 00:33:18.990 --> 00:33:21.680 align:middle line:90% and turn them into a set. 00:33:21.680 --> 00:33:23.750 align:middle line:84% And that reduces the number of options 00:33:23.750 --> 00:33:26.540 align:middle line:84% we have by k factorial on the left, and n 00:33:26.540 --> 00:33:29.940 align:middle line:84% minus k factorial on the right, the same way it did here. 00:33:29.940 --> 00:33:36.140 align:middle line:84% So this is n factorial divided by k factorial times 00:33:36.140 --> 00:33:39.830 align:middle line:90% n minus k factorial. 00:33:39.830 --> 00:33:42.120 align:middle line:90% This is a really useful formula. 00:33:42.120 --> 00:33:44.490 align:middle line:84% You're definitely going to want it on your cheat sheet. 00:33:44.490 --> 00:33:45.975 align:middle line:84% But in fact, hopefully by the time 00:33:45.975 --> 00:33:47.600 align:middle line:84% the quiz comes around next week, you'll 00:33:47.600 --> 00:33:50.700 align:middle line:84% be so used to this formula and so sick of-- 00:33:50.700 --> 00:33:53.000 align:middle line:84% I mean, you'll love it so much that you 00:33:53.000 --> 00:33:55.250 align:middle line:84% aren't you aren't going to need it on your cheat sheet 00:33:55.250 --> 00:33:58.550 align:middle line:90% because you're going to know it. 00:33:58.550 --> 00:34:02.450 align:middle line:84% And in fact, we have a special notation for it. 00:34:02.450 --> 00:34:06.680 align:middle line:84% We write it as n on the top, k on the bottom surrounded 00:34:06.680 --> 00:34:08.219 align:middle line:90% in parentheses. 00:34:08.219 --> 00:34:10.260 align:middle line:90% Notice it's not a fraction. 00:34:10.260 --> 00:34:11.760 align:middle line:90% We're not dividing n by k. 00:34:11.760 --> 00:34:14.320 align:middle line:84% So don't write n divided by k in parentheses. 00:34:14.320 --> 00:34:16.199 align:middle line:84% It's just n on top, k on the bottom, 00:34:16.199 --> 00:34:18.540 align:middle line:90% parentheses on the sides. 00:34:18.540 --> 00:34:25.080 align:middle line:84% And this is pronounced "n choose k." 00:34:25.080 --> 00:34:27.550 align:middle line:84% So n choose k is written like this. 00:34:27.550 --> 00:34:30.300 align:middle line:84% We have a formula for it like this. 00:34:30.300 --> 00:34:34.290 align:middle line:84% And it's pronounced "n choose k" because, well, 00:34:34.290 --> 00:34:39.600 align:middle line:84% we're choosing k things out of a possible set of n things. 00:34:39.600 --> 00:34:41.639 align:middle line:84% And like I said, really useful formula. 00:34:41.639 --> 00:34:45.540 align:middle line:84% If you're ordering pizza for dinner, there are 15 sides-- 00:34:45.540 --> 00:34:47.159 align:middle line:84% or sorry, 15 toppings that you can 00:34:47.159 --> 00:34:49.199 align:middle line:84% choose from to put on your pizza and you're 00:34:49.199 --> 00:34:52.860 align:middle line:84% supposed to choose three of them for some special promotion. 00:34:52.860 --> 00:34:54.940 align:middle line:84% How many ways are there to do that? 00:34:54.940 --> 00:35:03.460 align:middle line:84% If we've got 15 toppings, we want to choose three. 00:35:03.460 --> 00:35:05.670 align:middle line:84% How many ways are there to choose what your pizza is 00:35:05.670 --> 00:35:06.930 align:middle line:90% going to look like? 00:35:06.930 --> 00:35:09.190 align:middle line:84% We don't care about the order of the toppings. 00:35:09.190 --> 00:35:12.480 align:middle line:84% We just care about the set of three toppings that you pick. 00:35:12.480 --> 00:35:17.280 align:middle line:84% It's in the name 15, choose three. 00:35:17.280 --> 00:35:22.770 align:middle line:84% 15 choose three, also known as 15 factorial divided 00:35:22.770 --> 00:35:25.740 align:middle line:90% by 3 factorial 12 factorial. 00:35:25.740 --> 00:35:27.450 align:middle line:90% Yeah? 00:35:27.450 --> 00:35:29.490 align:middle line:90% Wonderful. 00:35:29.490 --> 00:35:32.070 align:middle line:84% If I need to select four volunteers out 00:35:32.070 --> 00:35:36.030 align:middle line:84% of this class of, let's say, 200 students, 00:35:36.030 --> 00:35:40.000 align:middle line:84% how many ways are there to pick a set of four volunteers? 00:35:40.000 --> 00:35:42.600 align:middle line:90% 200 choose four. 00:35:42.600 --> 00:35:44.693 align:middle line:90% If I flip a coin 100 times-- 00:35:44.693 --> 00:35:49.770 align:middle line:90% 00:35:49.770 --> 00:36:05.280 align:middle line:84% flip a coin 100 times, how many resulting sequences 00:36:05.280 --> 00:36:11.245 align:middle line:84% have 50 heads and 50 tails in some order? 00:36:11.245 --> 00:36:15.560 align:middle line:90% 00:36:15.560 --> 00:36:18.350 align:middle line:84% There are two times two times two times 00:36:18.350 --> 00:36:21.500 align:middle line:84% two 100 times number of options in total, 00:36:21.500 --> 00:36:23.390 align:middle line:90% two to the 100 options. 00:36:23.390 --> 00:36:27.380 align:middle line:84% How many of those results result in exactly 50 heads 00:36:27.380 --> 00:36:32.450 align:middle line:84% and exactly 50 tails, possibly intermixed? 00:36:32.450 --> 00:36:34.080 align:middle line:84% Well, if we remember from Tuesday, 00:36:34.080 --> 00:36:41.120 align:middle line:84% we had this bijection between binary sequences and subsets 00:36:41.120 --> 00:36:44.630 align:middle line:84% where we can turn a binary sequence into a subset 00:36:44.630 --> 00:36:47.910 align:middle line:84% by just taking all the ones and leaving out all the zeros. 00:36:47.910 --> 00:36:50.660 align:middle line:84% Take the indices of all the ones. 00:36:50.660 --> 00:36:52.320 align:middle line:84% And you can do that here as well. 00:36:52.320 --> 00:36:56.960 align:middle line:84% If we're looking at length 100 binary sequences 00:36:56.960 --> 00:37:02.430 align:middle line:84% with 50 ones and 50 zeros, where one is head and zero is tail. 00:37:02.430 --> 00:37:12.810 align:middle line:84% So number of length 100 binary sequences 00:37:12.810 --> 00:37:19.070 align:middle line:84% with 50 ones and 50 zeros from the bijection 00:37:19.070 --> 00:37:20.310 align:middle line:90% we talked about on Tuesday. 00:37:20.310 --> 00:37:28.250 align:middle line:84% This is in bijection with the number of size 00:37:28.250 --> 00:37:37.730 align:middle line:90% 50 subsets of one through 100. 00:37:37.730 --> 00:37:41.420 align:middle line:84% All we're doing is choosing 50 out of the 100 flips 00:37:41.420 --> 00:37:45.950 align:middle line:84% to be ones, to be heads, and the rest are going to be tails. 00:37:45.950 --> 00:37:48.722 align:middle line:84% So we're choosing 50 out of the 100. 00:37:48.722 --> 00:37:49.430 align:middle line:90% It's in the name. 00:37:49.430 --> 00:37:55.670 align:middle line:90% 100, choose 50. 00:37:55.670 --> 00:37:59.420 align:middle line:84% That's the number of ways to flip a coin 100 times 00:37:59.420 --> 00:38:01.650 align:middle line:84% and get exactly 50 heads and 50 tails. 00:38:01.650 --> 00:38:04.850 align:middle line:90% Did that make sense? 00:38:04.850 --> 00:38:05.870 align:middle line:90% Wonderful. 00:38:05.870 --> 00:38:09.240 align:middle line:84% Interestingly-- it's a little wet. 00:38:09.240 --> 00:38:10.160 align:middle line:90% Sorry. 00:38:10.160 --> 00:38:13.820 align:middle line:90% 100 choose 50 divided by-- 00:38:13.820 --> 00:38:14.750 align:middle line:90% what is it? 00:38:14.750 --> 00:38:16.590 align:middle line:90% 2 to the 100. 00:38:16.590 --> 00:38:21.050 align:middle line:84% So the ratio, the fraction of all results that have exactly 50 00:38:21.050 --> 00:38:23.870 align:middle line:84% heads and 50 tails out of the total number of results 00:38:23.870 --> 00:38:26.753 align:middle line:84% you can have, turns out this is about 8%. 00:38:26.753 --> 00:38:30.380 align:middle line:90% 00:38:30.380 --> 00:38:32.420 align:middle line:84% Which, to me, is a little counterintuitive. 00:38:32.420 --> 00:38:35.190 align:middle line:84% 50/50 has got to be the most common result. 00:38:35.190 --> 00:38:38.330 align:middle line:84% It's right in the middle, assuming we have a fair coin. 00:38:38.330 --> 00:38:42.440 align:middle line:84% And yet it's not actually all that likely. 00:38:42.440 --> 00:38:47.570 align:middle line:84% Turns out we're pretty likely to be very near 50%, 00:38:47.570 --> 00:38:50.210 align:middle line:84% but it's kind of hard to hit it exactly. 00:38:50.210 --> 00:38:52.650 align:middle line:84% If we were doing 1,000 flips instead of 100, 00:38:52.650 --> 00:38:54.540 align:middle line:90% I think it's closer to 1%. 00:38:54.540 --> 00:38:57.110 align:middle line:84% And it gets rarer and rarer as you go. 00:38:57.110 --> 00:39:00.230 align:middle line:84% But now we have the tools to count things like that. 00:39:00.230 --> 00:39:03.020 align:middle line:90% Questions about this n choose k. 00:39:03.020 --> 00:39:07.012 align:middle line:90% 00:39:07.012 --> 00:39:09.520 align:middle line:84% For reasons that we'll see next week, 00:39:09.520 --> 00:39:12.986 align:middle line:84% this n choose k is called a binomial coefficient. 00:39:12.986 --> 00:39:20.370 align:middle line:90% 00:39:20.370 --> 00:39:22.590 align:middle line:84% So binomial coefficient is the general name 00:39:22.590 --> 00:39:25.290 align:middle line:90% for these n choose k terms. 00:39:25.290 --> 00:39:28.330 align:middle line:90% Again, we'll see why on Tuesday. 00:39:28.330 --> 00:39:30.090 align:middle line:84% But if you see this term, know that that's 00:39:30.090 --> 00:39:33.030 align:middle line:90% what we're referring to. 00:39:33.030 --> 00:39:35.850 align:middle line:90% All right. 00:39:35.850 --> 00:39:38.340 align:middle line:90% Nice. 00:39:38.340 --> 00:39:40.860 align:middle line:84% Now the last technique I want to talk about, 00:39:40.860 --> 00:39:46.890 align:middle line:84% the last counting technique is extremely useful 00:39:46.890 --> 00:39:49.140 align:middle line:90% and extremely versatile. 00:39:49.140 --> 00:39:51.930 align:middle line:84% But that also means that there are 00:39:51.930 --> 00:39:54.510 align:middle line:84% lots of ways to apply it and also lots of ways 00:39:54.510 --> 00:39:57.090 align:middle line:90% to apply it incorrectly. 00:39:57.090 --> 00:39:59.350 align:middle line:84% So let's talk about how to do it. 00:39:59.350 --> 00:40:02.250 align:middle line:84% And more importantly, let's talk about what 00:40:02.250 --> 00:40:05.710 align:middle line:84% we need to make sure of before we know with confidence 00:40:05.710 --> 00:40:06.675 align:middle line:90% that we can apply it. 00:40:06.675 --> 00:40:12.970 align:middle line:90% 00:40:12.970 --> 00:40:16.690 align:middle line:84% And in fact, conveniently, it's built exactly 00:40:16.690 --> 00:40:19.472 align:middle line:84% on the generalized product rule, this next technique 00:40:19.472 --> 00:40:20.180 align:middle line:90% that we're using. 00:40:20.180 --> 00:40:23.040 align:middle line:84% In fact, I'm going to keep that up so we can look at it. 00:40:23.040 --> 00:40:36.700 align:middle line:90% 00:40:36.700 --> 00:40:39.850 align:middle line:84% And this technique, the book calls 00:40:39.850 --> 00:40:51.193 align:middle line:84% it counting with sequences, or counting with recipes. 00:40:51.193 --> 00:40:54.090 align:middle line:90% 00:40:54.090 --> 00:40:54.590 align:middle line:90% Excuse me. 00:40:54.590 --> 00:41:00.070 align:middle line:84% The general idea is if you have some set you're trying to count, 00:41:00.070 --> 00:41:02.800 align:middle line:84% well, it would be really convenient 00:41:02.800 --> 00:41:05.300 align:middle line:90% if you could build up your set-- 00:41:05.300 --> 00:41:08.710 align:middle line:84% build all the elements of your set via sequence 00:41:08.710 --> 00:41:11.590 align:middle line:84% of convenient choices where you always 00:41:11.590 --> 00:41:14.380 align:middle line:84% have the same number of choices every time. 00:41:14.380 --> 00:41:17.470 align:middle line:84% So then you can use the generalized product rule. 00:41:17.470 --> 00:41:19.450 align:middle line:84% So counting with sequences or recipes 00:41:19.450 --> 00:41:23.680 align:middle line:84% is just a way to construct your set with an easily digestible 00:41:23.680 --> 00:41:26.410 align:middle line:90% sequence of simple choices. 00:41:26.410 --> 00:41:28.990 align:middle line:84% And as long as you verify that that does indeed 00:41:28.990 --> 00:41:31.990 align:middle line:84% construct your set exactly, then you 00:41:31.990 --> 00:41:35.110 align:middle line:84% have your count with the generalized product rule. 00:41:35.110 --> 00:41:37.670 align:middle line:84% This will make a lot more sense when we start applying it. 00:41:37.670 --> 00:41:39.820 align:middle line:90% So let's start doing that. 00:41:39.820 --> 00:41:45.640 align:middle line:84% And the example we're going to use is counting poker hands. 00:41:45.640 --> 00:41:47.685 align:middle line:84% So we're going to use a standard deck of cards. 00:41:47.685 --> 00:41:56.710 align:middle line:90% 00:41:56.710 --> 00:42:00.690 align:middle line:84% A standard deck of cards has 13 of what are 00:42:00.690 --> 00:42:04.050 align:middle line:90% called ranks and four suits. 00:42:04.050 --> 00:42:07.800 align:middle line:84% So I'm told that some of you in the back 00:42:07.800 --> 00:42:10.890 align:middle line:84% can't see my comically oversized playing cards. 00:42:10.890 --> 00:42:12.620 align:middle line:90% So let's do better. 00:42:12.620 --> 00:42:15.510 align:middle line:90% 00:42:15.510 --> 00:42:17.110 align:middle line:90% So we have four suits. 00:42:17.110 --> 00:42:19.720 align:middle line:84% We have spades and hearts in my left hand. 00:42:19.720 --> 00:42:22.800 align:middle line:84% We have clubs and diamonds in my right hand. 00:42:22.800 --> 00:42:24.700 align:middle line:90% Those are the four suits. 00:42:24.700 --> 00:42:27.763 align:middle line:90% And then there are 13 ranks. 00:42:27.763 --> 00:42:30.990 align:middle line:90% 00:42:30.990 --> 00:42:32.620 align:middle line:90% No way this is going to work. 00:42:32.620 --> 00:42:35.260 align:middle line:90% That goes ace, two-- 00:42:35.260 --> 00:42:37.290 align:middle line:84% I'm not going to try to fan them out. 00:42:37.290 --> 00:42:38.310 align:middle line:90% Sorry. 00:42:38.310 --> 00:42:48.040 align:middle line:84% 3, 4, 5, 6, 7, 8, 9, 10, then Jack, then queen, then king. 00:42:48.040 --> 00:42:51.330 align:middle line:84% Those are the 13 different ranks. 00:42:51.330 --> 00:42:56.470 align:middle line:84% Sometimes in poker, the ace is the lowest of the numbers. 00:42:56.470 --> 00:42:58.740 align:middle line:84% Sometimes, it's the highest of the numbers. 00:42:58.740 --> 00:43:02.030 align:middle line:84% We're not really going to care about that so much. 00:43:02.030 --> 00:43:05.270 align:middle line:84% All we need to know is that there are four suits and 13 00:43:05.270 --> 00:43:09.200 align:middle line:84% ranks, and the deck consists of every combination 00:43:09.200 --> 00:43:10.320 align:middle line:90% of one of each. 00:43:10.320 --> 00:43:13.700 align:middle line:90% So 13 times 4 is 52 cards. 00:43:13.700 --> 00:43:18.320 align:middle line:90% So 13 ranks, four suits. 00:43:18.320 --> 00:43:21.803 align:middle line:84% These are called hearts, clubs, spades, and diamonds. 00:43:21.803 --> 00:43:24.740 align:middle line:90% 00:43:24.740 --> 00:43:29.840 align:middle line:84% And there are 52 cards corresponding to every way 00:43:29.840 --> 00:43:31.720 align:middle line:90% to choose one rank and one suit. 00:43:31.720 --> 00:43:35.330 align:middle line:90% 00:43:35.330 --> 00:43:37.770 align:middle line:84% That is the standard deck of cards. 00:43:37.770 --> 00:43:41.660 align:middle line:84% We're going to be assuming that for these next examples. 00:43:41.660 --> 00:43:45.170 align:middle line:90% Come to me, notes. 00:43:45.170 --> 00:43:51.560 align:middle line:84% And in poker, a poker hand is a set of five cards. 00:43:51.560 --> 00:44:01.650 align:middle line:84% So a poker hand is a subset of five cards out of the 52 00:44:01.650 --> 00:44:03.840 align:middle line:90% possible cards. 00:44:03.840 --> 00:44:07.290 align:middle line:84% And some subsets have more specialized properties 00:44:07.290 --> 00:44:10.510 align:middle line:84% that are less common, and so they're more powerful. 00:44:10.510 --> 00:44:14.670 align:middle line:84% And you get to beat your enemies with powerful hands. 00:44:14.670 --> 00:44:17.730 align:middle line:90% That's poker, right? 00:44:17.730 --> 00:44:20.140 align:middle line:84% So let's see some different kinds of poker hands. 00:44:20.140 --> 00:44:22.835 align:middle line:84% And the first one I want to look at is four of a kind. 00:44:22.835 --> 00:44:28.620 align:middle line:90% 00:44:28.620 --> 00:44:31.360 align:middle line:84% So a poker hand is called a four of a kind 00:44:31.360 --> 00:44:40.290 align:middle line:84% if you have all four cards of some rank and then 00:44:40.290 --> 00:44:42.660 align:middle line:90% some fifth card. 00:44:42.660 --> 00:44:44.850 align:middle line:84% So four of a kind, because you have all four 00:44:44.850 --> 00:44:47.520 align:middle line:90% options of the same rank. 00:44:47.520 --> 00:44:51.180 align:middle line:84% And we want to count how many four of a kind hands 00:44:51.180 --> 00:44:53.500 align:middle line:84% are there in a standard 52 deck-- 00:44:53.500 --> 00:44:56.290 align:middle line:90% 52-card deck. 00:44:56.290 --> 00:45:01.810 align:middle line:84% And we're going to do this by counting with a recipe here. 00:45:01.810 --> 00:45:05.870 align:middle line:84% And basically, I kind of already wrote the recipe, 00:45:05.870 --> 00:45:08.380 align:middle line:90% but let's do it carefully. 00:45:08.380 --> 00:45:16.030 align:middle line:84% Step one, I'm going to choose a rank R 00:45:16.030 --> 00:45:21.750 align:middle line:84% and take all four cards of that rank. 00:45:21.750 --> 00:45:25.150 align:middle line:90% 00:45:25.150 --> 00:45:32.350 align:middle line:84% And then I'm going to pick one other card. 00:45:32.350 --> 00:45:34.960 align:middle line:84% And putting these two steps together, 00:45:34.960 --> 00:45:41.560 align:middle line:84% we have now constructed a poker hand, a set of five cards. 00:45:41.560 --> 00:45:44.800 align:middle line:84% And the claim is that this recipe 00:45:44.800 --> 00:45:48.165 align:middle line:84% is going to help us count the set of four of a kind hands. 00:45:48.165 --> 00:45:51.220 align:middle line:90% 00:45:51.220 --> 00:45:52.750 align:middle line:90% Great. 00:45:52.750 --> 00:45:56.200 align:middle line:84% We can be a little more precise with this. 00:45:56.200 --> 00:45:59.260 align:middle line:90% What we've really done-- 00:45:59.260 --> 00:46:00.370 align:middle line:90% excuse me. 00:46:00.370 --> 00:46:01.870 align:middle line:90% So first, we've set up a set. 00:46:01.870 --> 00:46:06.760 align:middle line:84% I'm going to call it A. It's going to be the set of R 00:46:06.760 --> 00:46:21.340 align:middle line:84% and C, where R is a rank, C is a card with rank not equal to R. 00:46:21.340 --> 00:46:23.445 align:middle line:90% So this is following our recipe. 00:46:23.445 --> 00:46:25.070 align:middle line:84% We choose a rank for the four of a kind 00:46:25.070 --> 00:46:27.695 align:middle line:84% and then we choose a fifth card, which can't be one of the four 00:46:27.695 --> 00:46:29.890 align:middle line:90% we've already picked. 00:46:29.890 --> 00:46:37.683 align:middle line:84% And B is the set of four of a kind hands. 00:46:37.683 --> 00:46:40.510 align:middle line:90% 00:46:40.510 --> 00:46:44.090 align:middle line:84% We have the set A of pairs, of sequences. 00:46:44.090 --> 00:46:47.920 align:middle line:84% This is where counting with sequences comes in. 00:46:47.920 --> 00:46:51.790 align:middle line:84% But B is really the set we're trying to count. 00:46:51.790 --> 00:47:01.970 align:middle line:84% And we've set up a map from A to B. 00:47:01.970 --> 00:47:05.270 align:middle line:84% We set up some function, f, from A 00:47:05.270 --> 00:47:10.040 align:middle line:84% to B, which takes a rank and some other card 00:47:10.040 --> 00:47:16.440 align:middle line:84% and turns it into the hand rank of hearts, rank of diamonds, 00:47:16.440 --> 00:47:24.710 align:middle line:84% rank of spades, rank of clubs, and then this other card C. 00:47:24.710 --> 00:47:27.590 align:middle line:84% Does that make sense, how this description of a recipe 00:47:27.590 --> 00:47:30.865 align:middle line:84% turns into this set, this set, and this map between them? 00:47:30.865 --> 00:47:34.980 align:middle line:90% 00:47:34.980 --> 00:47:35.480 align:middle line:90% Wonderful. 00:47:35.480 --> 00:47:38.300 align:middle line:90% No objections, so I'll continue. 00:47:38.300 --> 00:47:44.670 align:middle line:84% And now the claim f is a bijection. 00:47:44.670 --> 00:47:48.200 align:middle line:90% 00:47:48.200 --> 00:47:50.160 align:middle line:84% This is where counting with recipes-- 00:47:50.160 --> 00:47:51.960 align:middle line:90% this is what the technique is. 00:47:51.960 --> 00:47:54.370 align:middle line:84% We want to construct some set of sequences 00:47:54.370 --> 00:47:57.830 align:middle line:84% where we're making a sequence of choices in order 00:47:57.830 --> 00:47:59.650 align:middle line:84% and we want to prove that the resulting 00:47:59.650 --> 00:48:05.080 align:middle line:84% set is in bijection with the set we're actually trying to count. 00:48:05.080 --> 00:48:07.040 align:middle line:84% And as long as we can verify that bijection, 00:48:07.040 --> 00:48:10.780 align:middle line:84% then we know the size of B, because we know the size of A 00:48:10.780 --> 00:48:13.270 align:middle line:90% with generalized product rule. 00:48:13.270 --> 00:48:17.300 align:middle line:84% So we need to verify f is a bijection. 00:48:17.300 --> 00:48:18.940 align:middle line:84% And then we need to count the size of A 00:48:18.940 --> 00:48:21.610 align:middle line:84% in order to figure out the size of B. 00:48:21.610 --> 00:48:25.150 align:middle line:84% So what does it mean for f to be a bijection? 00:48:25.150 --> 00:48:34.630 align:middle line:84% Well, we need f to be a function, total, surjective, 00:48:34.630 --> 00:48:36.326 align:middle line:90% and injective. 00:48:36.326 --> 00:48:44.170 align:middle line:90% 00:48:44.170 --> 00:48:47.450 align:middle line:84% So what does it mean for f to be a function? 00:48:47.450 --> 00:48:50.820 align:middle line:84% It means for every input, there's, at most, one output. 00:48:50.820 --> 00:48:54.150 align:middle line:84% And that's usually built into the process. 00:48:54.150 --> 00:48:56.310 align:middle line:84% We have some deterministic way of turning 00:48:56.310 --> 00:48:59.740 align:middle line:84% this pair into this hand and there's not ambiguity. 00:48:59.740 --> 00:49:02.140 align:middle line:84% This pair can't turn into seven different hands. 00:49:02.140 --> 00:49:04.620 align:middle line:84% Usually, we get the function side for free 00:49:04.620 --> 00:49:07.530 align:middle line:84% and it's not really worth commenting on much. 00:49:07.530 --> 00:49:09.810 align:middle line:90% What does total mean? 00:49:09.810 --> 00:49:16.383 align:middle line:84% Total means that every pair turns into at least one hand. 00:49:16.383 --> 00:49:17.800 align:middle line:84% And since we know it's a function, 00:49:17.800 --> 00:49:19.990 align:middle line:84% every pair turns into exactly one hand. 00:49:19.990 --> 00:49:25.530 align:middle line:84% Specifically, it turns into a hand in the set we care about. 00:49:25.530 --> 00:49:34.140 align:middle line:84% So we have our first meaningful question, is f of R, C, a four 00:49:34.140 --> 00:49:45.280 align:middle line:84% of a kind for every way we have to fill out our recipe? 00:49:45.280 --> 00:49:47.880 align:middle line:84% So is everything we construct-- are all of the hands 00:49:47.880 --> 00:49:52.250 align:middle line:84% that we construct really four of a kind hands, 00:49:52.250 --> 00:49:54.920 align:middle line:84% or are we constructing some extraneous stuff? 00:49:54.920 --> 00:49:56.630 align:middle line:84% If we're building extraneous stuff, 00:49:56.630 --> 00:49:58.213 align:middle line:84% then we're not counting the right set, 00:49:58.213 --> 00:50:00.130 align:middle line:84% because we're not building the right set. 00:50:00.130 --> 00:50:02.950 align:middle line:90% So first meaningful question. 00:50:02.950 --> 00:50:05.560 align:middle line:84% For every way we have to fill the recipe, 00:50:05.560 --> 00:50:08.680 align:middle line:84% is the hand that we've constructed a hand 00:50:08.680 --> 00:50:11.080 align:middle line:90% that we're trying to count? 00:50:11.080 --> 00:50:14.430 align:middle line:84% And in this case, well, we can see that yes, yes, it 00:50:14.430 --> 00:50:18.490 align:middle line:84% is because we have all four cards of the same rank. 00:50:18.490 --> 00:50:22.480 align:middle line:84% And we have a fifth card that is not any of these cards 00:50:22.480 --> 00:50:25.040 align:middle line:84% because it has rank that's different from R. 00:50:25.040 --> 00:50:27.670 align:middle line:84% So this really is a size five subset with four cards 00:50:27.670 --> 00:50:28.490 align:middle line:90% of the same rank. 00:50:28.490 --> 00:50:32.050 align:middle line:90% So in this case, yes. 00:50:32.050 --> 00:50:34.750 align:middle line:90% Great. 00:50:34.750 --> 00:50:37.900 align:middle line:90% We need f to be surjective. 00:50:37.900 --> 00:50:39.830 align:middle line:84% What does it mean to be surjective? 00:50:39.830 --> 00:50:44.530 align:middle line:84% It means every four of a kind hand, everything in B, 00:50:44.530 --> 00:50:57.130 align:middle line:84% comes from something in A. Can every four of a kind hand 00:50:57.130 --> 00:50:58.020 align:middle line:90% be built? 00:50:58.020 --> 00:51:02.770 align:middle line:90% 00:51:02.770 --> 00:51:04.930 align:middle line:84% So can we really construct everything 00:51:04.930 --> 00:51:07.300 align:middle line:90% that we're trying to count? 00:51:07.300 --> 00:51:17.570 align:middle line:84% In this example-- let's look at an example. 00:51:17.570 --> 00:51:21.170 align:middle line:84% If I have the four of a kind hand that has a three of hearts, 00:51:21.170 --> 00:51:25.900 align:middle line:84% the seven of hearts, the three of diamonds, the three of clubs, 00:51:25.900 --> 00:51:31.700 align:middle line:84% and the three of spades, how do I build that from my recipe? 00:51:31.700 --> 00:51:35.680 align:middle line:84% How do I get that as an output of my function f? 00:51:35.680 --> 00:51:37.665 align:middle line:90% What is R and what is C? 00:51:37.665 --> 00:51:50.420 align:middle line:90% 00:51:50.420 --> 00:51:52.316 align:middle line:90% Yep. 00:51:52.316 --> 00:51:56.162 align:middle line:90% AUDIENCE: R is 3, and C is 7. 00:51:56.162 --> 00:51:56.870 align:middle line:90% ZACHARY ABEL: OK. 00:51:56.870 --> 00:52:01.940 align:middle line:84% R is three and C is the seven of hearts. 00:52:01.940 --> 00:52:03.680 align:middle line:90% Oops. 00:52:03.680 --> 00:52:05.220 align:middle line:90% That's exactly right. 00:52:05.220 --> 00:52:06.950 align:middle line:84% The rank R is supposed to tell us 00:52:06.950 --> 00:52:08.910 align:middle line:90% the rank that we have four of. 00:52:08.910 --> 00:52:10.020 align:middle line:90% We have four 3's. 00:52:10.020 --> 00:52:11.310 align:middle line:90% So the rank is 3. 00:52:11.310 --> 00:52:14.450 align:middle line:84% The card C is supposed to be the fifth extraneous card 00:52:14.450 --> 00:52:17.360 align:middle line:84% in the hand that isn't part of the four of a kind. 00:52:17.360 --> 00:52:20.060 align:middle line:84% And that's the 7 of hearts in this example. 00:52:20.060 --> 00:52:21.570 align:middle line:90% Was there any ambiguity? 00:52:21.570 --> 00:52:24.200 align:middle line:90% 00:52:24.200 --> 00:52:24.750 align:middle line:90% Sorry. 00:52:24.750 --> 00:52:27.080 align:middle line:84% More importantly, will we always be able to do this? 00:52:27.080 --> 00:52:29.900 align:middle line:84% For every four of a kind hand, will we always 00:52:29.900 --> 00:52:33.260 align:middle line:84% be able to pick out a rank and pick out the fifth card? 00:52:33.260 --> 00:52:34.960 align:middle line:90% Yeah. 00:52:34.960 --> 00:52:35.460 align:middle line:90% Yeah. 00:52:35.460 --> 00:52:37.640 align:middle line:90% So we are surjective. 00:52:37.640 --> 00:52:43.190 align:middle line:84% We can build every hand that we care to build. 00:52:43.190 --> 00:52:50.040 align:middle line:84% The last thing we need to check is, is our function injective? 00:52:50.040 --> 00:52:55.500 align:middle line:84% And what it means for f to be injective is 00:52:55.500 --> 00:53:11.580 align:middle line:84% can every four of a kind hand be built in exactly one way? 00:53:11.580 --> 00:53:15.060 align:middle line:84% Exactly one way is what it means to be surjective and injective 00:53:15.060 --> 00:53:17.940 align:middle line:90% combined. 00:53:17.940 --> 00:53:21.400 align:middle line:84% So for every four of a kind hand, 00:53:21.400 --> 00:53:25.620 align:middle line:84% is there only one way to build it with our recipe? 00:53:25.620 --> 00:53:28.410 align:middle line:84% And in this case, yeah, for kind of the same reason. 00:53:28.410 --> 00:53:30.930 align:middle line:84% We know which rank we have four of. 00:53:30.930 --> 00:53:33.258 align:middle line:84% We know which card is the other card that 00:53:33.258 --> 00:53:35.800 align:middle line:84% doesn't match that rank, and there's only one way to do that. 00:53:35.800 --> 00:53:38.050 align:middle line:84% In fact, there's exactly one way to do that. 00:53:38.050 --> 00:53:40.920 align:middle line:84% So we have built everything in our set 00:53:40.920 --> 00:53:44.670 align:middle line:90% exactly once with our recipe. 00:53:44.670 --> 00:53:50.070 align:middle line:84% So in this case, yes, which means it is injective. 00:53:50.070 --> 00:53:55.450 align:middle line:84% And now we've decided, therefore f is a bijection. 00:53:55.450 --> 00:53:59.730 align:middle line:90% 00:53:59.730 --> 00:54:01.560 align:middle line:90% Are we done? 00:54:01.560 --> 00:54:02.110 align:middle line:90% Wait. 00:54:02.110 --> 00:54:04.140 align:middle line:90% What were we even trying to do? 00:54:04.140 --> 00:54:04.750 align:middle line:90% Oh yeah. 00:54:04.750 --> 00:54:07.000 align:middle line:84% We were trying to count the number of four of a kinds. 00:54:07.000 --> 00:54:09.420 align:middle line:90% 00:54:09.420 --> 00:54:12.450 align:middle line:84% We did one of the two important steps. 00:54:12.450 --> 00:54:17.310 align:middle line:84% We showed that A is in bijection with B. 00:54:17.310 --> 00:54:20.400 align:middle line:84% But now we need the actual number, the actual size of B. 00:54:20.400 --> 00:54:22.050 align:middle line:90% We need the count. 00:54:22.050 --> 00:54:23.910 align:middle line:84% And if we want to count the size of B, 00:54:23.910 --> 00:54:28.650 align:middle line:84% now we know it's enough to count the size of A. 00:54:28.650 --> 00:54:37.040 align:middle line:84% So we can use this time the generalized product rule. 00:54:37.040 --> 00:54:43.540 align:middle line:90% 00:54:43.540 --> 00:54:45.940 align:middle line:90% So size of a. 00:54:45.940 --> 00:54:49.720 align:middle line:84% Well remember, we're choosing an R and we're choosing a C. 00:54:49.720 --> 00:54:52.965 align:middle line:84% So how many options do we have for our rank R? 00:54:52.965 --> 00:54:56.168 align:middle line:90% 00:54:56.168 --> 00:54:56.710 align:middle line:90% AUDIENCE: 13. 00:54:56.710 --> 00:54:57.418 align:middle line:90% ZACHARY ABEL: 13. 00:54:57.418 --> 00:54:59.330 align:middle line:84% There are 13 ranks ace through king. 00:54:59.330 --> 00:55:01.720 align:middle line:84% We can choose any one of them freely. 00:55:01.720 --> 00:55:05.140 align:middle line:84% How many options do we have for this last card, C, 00:55:05.140 --> 00:55:06.645 align:middle line:90% that doesn't have this rank? 00:55:06.645 --> 00:55:11.240 align:middle line:90% 00:55:11.240 --> 00:55:12.160 align:middle line:90% AUDIENCE: 48. 00:55:12.160 --> 00:55:13.390 align:middle line:90% ZACHARY ABEL: 48? 00:55:13.390 --> 00:55:14.270 align:middle line:90% Perfect. 00:55:14.270 --> 00:55:15.650 align:middle line:90% There are 52 cards in the deck. 00:55:15.650 --> 00:55:18.280 align:middle line:84% We've already chosen four of them of this rank. 00:55:18.280 --> 00:55:20.660 align:middle line:90% There are 48 options left. 00:55:20.660 --> 00:55:28.150 align:middle line:84% So by the generalized product rule, the size of a 00:55:28.150 --> 00:55:29.520 align:middle line:90% is 13 times 48. 00:55:29.520 --> 00:55:36.220 align:middle line:90% 00:55:36.220 --> 00:55:40.820 align:middle line:84% So that's the other thing our recipe needs to satisfy. 00:55:40.820 --> 00:55:44.130 align:middle line:84% It needs to form a bijection with the set we care about, 00:55:44.130 --> 00:55:46.100 align:middle line:84% but it also needs to be easy to count 00:55:46.100 --> 00:55:48.530 align:middle line:84% with the generalized product rule. 00:55:48.530 --> 00:55:50.430 align:middle line:90% And that's the strategy. 00:55:50.430 --> 00:55:52.100 align:middle line:90% Break down your set. 00:55:52.100 --> 00:55:53.960 align:middle line:84% You can think of it as building up 00:55:53.960 --> 00:55:57.380 align:middle line:84% your set via a sequence of simple choices. 00:55:57.380 --> 00:55:59.840 align:middle line:90% Make sure it's a bijection. 00:55:59.840 --> 00:56:02.720 align:middle line:84% And then count it with generalized product rule. 00:56:02.720 --> 00:56:03.930 align:middle line:90% Let's do that again. 00:56:03.930 --> 00:56:05.780 align:middle line:90% Let's look at another example. 00:56:05.780 --> 00:56:14.533 align:middle line:84% This time, I want the number of hands with all four suits. 00:56:14.533 --> 00:56:17.450 align:middle line:90% 00:56:17.450 --> 00:56:20.720 align:middle line:84% How many hands have all four suits? 00:56:20.720 --> 00:56:22.260 align:middle line:90% These are useless in poker. 00:56:22.260 --> 00:56:24.770 align:middle line:84% This doesn't actually get you anything, 00:56:24.770 --> 00:56:27.813 align:middle line:84% unless it has some other coincidences as well. 00:56:27.813 --> 00:56:28.730 align:middle line:90% But it's fun to count. 00:56:28.730 --> 00:56:30.930 align:middle line:84% How many hands have all four suits? 00:56:30.930 --> 00:56:34.340 align:middle line:84% And let's see if we can break this down into a recipe. 00:56:34.340 --> 00:56:34.890 align:middle line:90% All right. 00:56:34.890 --> 00:56:35.643 align:middle line:90% Here's my recipe. 00:56:35.643 --> 00:56:39.140 align:middle line:90% 00:56:39.140 --> 00:56:41.425 align:middle line:90% Choose a heart card. 00:56:41.425 --> 00:56:46.400 align:middle line:90% 00:56:46.400 --> 00:56:48.110 align:middle line:90% Choose-- I'm sorry. 00:56:48.110 --> 00:56:50.340 align:middle line:84% My notes have-- it was a club card first. 00:56:50.340 --> 00:56:51.550 align:middle line:90% Then choose a heart card. 00:56:51.550 --> 00:56:55.460 align:middle line:90% 00:56:55.460 --> 00:56:56.920 align:middle line:90% Then choose a spade card. 00:56:56.920 --> 00:57:01.760 align:middle line:90% 00:57:01.760 --> 00:57:04.670 align:middle line:90% Choose a diamond card. 00:57:04.670 --> 00:57:08.290 align:middle line:90% And then choose a fifth card. 00:57:08.290 --> 00:57:11.630 align:middle line:90% 00:57:11.630 --> 00:57:12.780 align:middle line:90% Here's my recipe. 00:57:12.780 --> 00:57:15.342 align:middle line:84% Choose a card that has clubs, a card that 00:57:15.342 --> 00:57:16.800 align:middle line:84% has hearts, a card that has spades, 00:57:16.800 --> 00:57:19.360 align:middle line:84% a card that has diamonds, and then whatever other card I 00:57:19.360 --> 00:57:22.190 align:middle line:90% want to fill the last spot. 00:57:22.190 --> 00:57:26.610 align:middle line:84% And let's see if this is a valid recipe for this problem. 00:57:26.610 --> 00:57:28.970 align:middle line:90% And we can boil that down. 00:57:28.970 --> 00:57:32.300 align:middle line:84% We don't always have to think, OK, injective, surjective, 00:57:32.300 --> 00:57:33.270 align:middle line:90% all that fun stuff. 00:57:33.270 --> 00:57:37.190 align:middle line:84% We can boil it down into these bullet points, which we can even 00:57:37.190 --> 00:57:39.270 align:middle line:90% compress a little further. 00:57:39.270 --> 00:57:54.630 align:middle line:84% So first of all, is every hand constructed using our recipe 00:57:54.630 --> 00:57:58.500 align:middle line:90% actually in our set? 00:57:58.500 --> 00:58:03.990 align:middle line:84% So does every hand we construct really have all four suits? 00:58:03.990 --> 00:58:09.365 align:middle line:90% And finally, does-- sorry. 00:58:09.365 --> 00:58:16.800 align:middle line:90% 00:58:16.800 --> 00:58:29.945 align:middle line:84% Can every hand in our set be constructed in exactly one way? 00:58:29.945 --> 00:58:32.910 align:middle line:90% 00:58:32.910 --> 00:58:35.530 align:middle line:84% These are the two questions we need to ask. 00:58:35.530 --> 00:58:38.080 align:middle line:84% Can we build everything we care about 00:58:38.080 --> 00:58:43.065 align:middle line:84% and can we build everything we care about in exactly one way? 00:58:43.065 --> 00:58:45.700 align:middle line:90% 00:58:45.700 --> 00:58:47.690 align:middle line:84% And if those two things are true, 00:58:47.690 --> 00:58:52.210 align:middle line:84% then our map is a bijection and our recipe is valid. 00:58:52.210 --> 00:58:55.120 align:middle line:84% But these are maybe easier to think about and remember 00:58:55.120 --> 00:58:57.680 align:middle line:84% than going through all that bijection rigmarole. 00:58:57.680 --> 00:58:59.900 align:middle line:84% Just remember that it's a bijection underneath. 00:58:59.900 --> 00:59:00.950 align:middle line:90% That's what we're doing. 00:59:00.950 --> 00:59:02.158 align:middle line:90% That's what the shepherds do. 00:59:02.158 --> 00:59:04.180 align:middle line:90% They count with bijections. 00:59:04.180 --> 00:59:08.290 align:middle line:90% So let's answer these questions. 00:59:08.290 --> 00:59:12.280 align:middle line:84% Is every hand constructed with this recipe actually 00:59:12.280 --> 00:59:13.850 align:middle line:90% a hand that has all four sets-- 00:59:13.850 --> 00:59:15.880 align:middle line:90% all four suits? 00:59:15.880 --> 00:59:17.050 align:middle line:90% Yeah? 00:59:17.050 --> 00:59:18.620 align:middle line:90% Yeah, absolutely. 00:59:18.620 --> 00:59:20.112 align:middle line:84% These first four cards are always 00:59:20.112 --> 00:59:21.320 align:middle line:90% going to have all four suits. 00:59:21.320 --> 00:59:24.350 align:middle line:84% And the fifth card doesn't really help or hurt us. 00:59:24.350 --> 00:59:26.690 align:middle line:84% So the answer to this first question is yes. 00:59:26.690 --> 00:59:28.750 align:middle line:90% Yes, it does. 00:59:28.750 --> 00:59:32.590 align:middle line:84% Can every hand in our set-- can every hand with all four suits 00:59:32.590 --> 00:59:38.950 align:middle line:84% be constructed via recipe in exactly one way? 00:59:38.950 --> 00:59:40.850 align:middle line:84% And let's look at an example for that. 00:59:40.850 --> 00:59:42.030 align:middle line:90% Examples are really helpful. 00:59:42.030 --> 00:59:52.480 align:middle line:90% 00:59:52.480 --> 00:59:53.710 align:middle line:90% So here's my hand. 00:59:53.710 --> 01:00:00.340 align:middle line:84% I'm going to look at the six of diamonds, the queen of hearts, 01:00:00.340 --> 01:00:07.870 align:middle line:84% the ace of spades, the two of clubs, and the Jack of spades. 01:00:07.870 --> 01:00:11.350 align:middle line:84% Let's say we're trying to build this particular hand using 01:00:11.350 --> 01:00:13.270 align:middle line:90% our recipe. 01:00:13.270 --> 01:00:15.050 align:middle line:84% Well, what does it tell us to do? 01:00:15.050 --> 01:00:16.850 align:middle line:90% First, choose a club card. 01:00:16.850 --> 01:00:17.510 align:middle line:90% All right. 01:00:17.510 --> 01:00:19.250 align:middle line:84% I'm going to choose two of clubs. 01:00:19.250 --> 01:00:20.570 align:middle line:90% Then choose a heart card. 01:00:20.570 --> 01:00:21.070 align:middle line:90% Great. 01:00:21.070 --> 01:00:24.280 align:middle line:84% I'm going to choose the queen of hearts. 01:00:24.280 --> 01:00:26.530 align:middle line:90% Then choose a spade card. 01:00:26.530 --> 01:00:28.720 align:middle line:90% Great, ace of spades. 01:00:28.720 --> 01:00:32.920 align:middle line:84% Then choose a diamond card, six of diamonds. 01:00:32.920 --> 01:00:38.510 align:middle line:84% And then my last card, whatever that is, it's Jack of spades. 01:00:38.510 --> 01:00:42.680 align:middle line:84% So these are the choices I make in the five steps of my recipe. 01:00:42.680 --> 01:00:44.620 align:middle line:90% I indeed have built this hand. 01:00:44.620 --> 01:00:47.240 align:middle line:90% 01:00:47.240 --> 01:00:50.570 align:middle line:84% Can I do that in exactly one way? 01:00:50.570 --> 01:00:52.970 align:middle line:84% Is there only one way to fill out this recipe in order 01:00:52.970 --> 01:00:53.845 align:middle line:90% to get this hand? 01:00:53.845 --> 01:00:58.880 align:middle line:90% 01:00:58.880 --> 01:01:00.210 align:middle line:90% Can someone find another? 01:01:00.210 --> 01:01:01.055 align:middle line:90% Yep. 01:01:01.055 --> 01:01:04.250 align:middle line:84% AUDIENCE: You could replace the A with the Jack. 01:01:04.250 --> 01:01:07.290 align:middle line:84% ZACHARY ABEL: Yeah, the ace and the jack, the two spade cards, 01:01:07.290 --> 01:01:09.800 align:middle line:90% those could have swapped places. 01:01:09.800 --> 01:01:13.020 align:middle line:84% In step three when I was told to pick a spade card, 01:01:13.020 --> 01:01:15.090 align:middle line:84% I could have picked the Jack instead. 01:01:15.090 --> 01:01:18.410 align:middle line:84% And then the ace would be my fifth extraneous card 01:01:18.410 --> 01:01:20.300 align:middle line:90% at the end. 01:01:20.300 --> 01:01:22.520 align:middle line:84% So there are, in fact, two different ways 01:01:22.520 --> 01:01:24.710 align:middle line:84% of filling out my recipe in order 01:01:24.710 --> 01:01:27.410 align:middle line:90% to get this particular hand. 01:01:27.410 --> 01:01:28.830 align:middle line:90% So it's not a bijection. 01:01:28.830 --> 01:01:31.020 align:middle line:84% We have failed our second condition. 01:01:31.020 --> 01:01:35.840 align:middle line:84% This recipe does not work with this method. 01:01:35.840 --> 01:01:37.950 align:middle line:90% Sometimes, you can salvage it. 01:01:37.950 --> 01:01:40.290 align:middle line:84% In this particular example, you can salvage it. 01:01:40.290 --> 01:01:47.870 align:middle line:84% It turns out this map that we've constructed 01:01:47.870 --> 01:01:52.160 align:middle line:84% is two to one, because there are always exactly two ways 01:01:52.160 --> 01:01:54.260 align:middle line:90% to fill out the recipe. 01:01:54.260 --> 01:01:56.820 align:middle line:84% Whichever card you have in the last spot, 01:01:56.820 --> 01:01:59.600 align:middle line:84% it can potentially swap places with its partner 01:01:59.600 --> 01:02:01.170 align:middle line:90% that has its same suit. 01:02:01.170 --> 01:02:04.130 align:middle line:84% But there are always only two ways to do that. 01:02:04.130 --> 01:02:06.240 align:middle line:84% So instead of constructing a bijection, 01:02:06.240 --> 01:02:09.020 align:middle line:84% we've constructed a two to one map, 01:02:09.020 --> 01:02:12.320 align:middle line:84% which means we can still count the number of ways 01:02:12.320 --> 01:02:15.950 align:middle line:84% of filling out the recipe and then just divide it by two. 01:02:15.950 --> 01:02:17.170 align:middle line:90% So let's do that. 01:02:17.170 --> 01:02:20.360 align:middle line:90% 01:02:20.360 --> 01:02:22.397 align:middle line:84% Number of ways to fill out this recipe. 01:02:22.397 --> 01:02:23.980 align:middle line:84% Actually, let's just do it right here. 01:02:23.980 --> 01:02:26.720 align:middle line:90% 01:02:26.720 --> 01:02:32.160 align:middle line:84% How many ways are there to choose a card that has clubs? 01:02:32.160 --> 01:02:35.370 align:middle line:90% Just call it out. 01:02:35.370 --> 01:02:38.550 align:middle line:84% How many club cards are there in a deck? 01:02:38.550 --> 01:02:39.910 align:middle line:90% 13, yeah. 01:02:39.910 --> 01:02:41.080 align:middle line:90% I think I heard that. 01:02:41.080 --> 01:02:43.290 align:middle line:90% Let's pretend I did. 01:02:43.290 --> 01:02:45.208 align:middle line:84% How many heart cards are in a deck? 01:02:45.208 --> 01:02:45.750 align:middle line:90% AUDIENCE: 13. 01:02:45.750 --> 01:02:47.310 align:middle line:90% ZACHARY ABEL: 13. 01:02:47.310 --> 01:02:48.930 align:middle line:90% How many spades? 01:02:48.930 --> 01:02:49.590 align:middle line:90% AUDIENCE: 13. 01:02:49.590 --> 01:02:50.405 align:middle line:90% ZACHARY ABEL: How many diamonds? 01:02:50.405 --> 01:02:50.860 align:middle line:90% AUDIENCE: 13. 01:02:50.860 --> 01:02:51.568 align:middle line:90% ZACHARY ABEL: 13. 01:02:51.568 --> 01:02:54.480 align:middle line:84% How many choices do I have for my last card? 01:02:54.480 --> 01:02:57.360 align:middle line:90% 48 again. 01:02:57.360 --> 01:02:58.950 align:middle line:84% I'm allowed to choose any card that 01:02:58.950 --> 01:03:01.920 align:middle line:84% isn't one of the four I've already picked, 01:03:01.920 --> 01:03:04.920 align:middle line:84% which means the number of ways to fill out this recipe 01:03:04.920 --> 01:03:09.300 align:middle line:84% is 13 times 13 times 13 times 13 times 48. 01:03:09.300 --> 01:03:16.230 align:middle line:84% So our answer is 13 to the fourth times 48 01:03:16.230 --> 01:03:21.360 align:middle line:84% divided by 2, because we decided it was a two to one map. 01:03:21.360 --> 01:03:26.730 align:middle line:84% So this is the number of hands that have all four suits. 01:03:26.730 --> 01:03:28.740 align:middle line:90% Did that make sense? 01:03:28.740 --> 01:03:31.720 align:middle line:84% This is a perfectly valid way to do this. 01:03:31.720 --> 01:03:35.560 align:middle line:84% This counting with sequences technique 01:03:35.560 --> 01:03:37.690 align:middle line:84% generalizes very nicely to this k 01:03:37.690 --> 01:03:42.020 align:middle line:84% to 1 mapping instead of 1 to 1 mapping, a bijection. 01:03:42.020 --> 01:03:44.290 align:middle line:84% Bijections are maybe more elegant, 01:03:44.290 --> 01:03:46.370 align:middle line:84% or easier to think about, if you can make it. 01:03:46.370 --> 01:03:49.100 align:middle line:84% But if it's 2 to 1 instead, that's great. 01:03:49.100 --> 01:03:51.220 align:middle line:90% Just divide by 2. 01:03:51.220 --> 01:03:55.450 align:middle line:84% Just so we can see some different strategies, 01:03:55.450 --> 01:03:58.690 align:middle line:84% here is a way that we could have done that 01:03:58.690 --> 01:04:01.185 align:middle line:84% with a bijection instead of needing a 2 to 1 map. 01:04:01.185 --> 01:04:09.280 align:middle line:90% 01:04:09.280 --> 01:04:11.930 align:middle line:84% So a different recipe is the following. 01:04:11.930 --> 01:04:24.783 align:middle line:84% So first, choose a suit S that will have two cards. 01:04:24.783 --> 01:04:28.120 align:middle line:90% 01:04:28.120 --> 01:04:31.940 align:middle line:84% So in every hand that has all four suits, 01:04:31.940 --> 01:04:33.730 align:middle line:84% there's always one suit that has two cards 01:04:33.730 --> 01:04:35.870 align:middle line:84% and the other three suits have one card. 01:04:35.870 --> 01:04:39.210 align:middle line:84% So first, we're going to choose which suit has two cards. 01:04:39.210 --> 01:04:42.850 align:middle line:90% 01:04:42.850 --> 01:04:51.885 align:middle line:84% Call the other three suits T1, T2, T3 in alphabetical order. 01:04:51.885 --> 01:04:57.490 align:middle line:90% 01:04:57.490 --> 01:04:59.560 align:middle line:84% So once I've chosen which suit I'm 01:04:59.560 --> 01:05:01.960 align:middle line:84% going to single out for the pair, 01:05:01.960 --> 01:05:04.310 align:middle line:84% I've just given names to the other three suits. 01:05:04.310 --> 01:05:07.160 align:middle line:84% And it's in alphabetical order, so there's no ambiguity there. 01:05:07.160 --> 01:05:12.890 align:middle line:84% So now my four suits are S and T1, T2, T3. 01:05:12.890 --> 01:05:27.230 align:middle line:84% Next thing I'm going to do, choose two cards with suit S. 01:05:27.230 --> 01:05:32.160 align:middle line:90% Choose one card with suit T1. 01:05:32.160 --> 01:05:33.990 align:middle line:90% Then choose one with suit T2. 01:05:33.990 --> 01:05:35.398 align:middle line:90% Then choose one with suit T3. 01:05:35.398 --> 01:05:36.273 align:middle line:90% And that's my recipe. 01:05:36.273 --> 01:05:43.010 align:middle line:90% 01:05:43.010 --> 01:05:46.050 align:middle line:84% And it turns out that this really does give a bijection. 01:05:46.050 --> 01:05:48.170 align:middle line:84% This is a one to one map between ways 01:05:48.170 --> 01:05:53.330 align:middle line:84% of filling out this recipe and hands that have all four suits. 01:05:53.330 --> 01:05:56.180 align:middle line:84% Because that ambiguity of which of these two cards 01:05:56.180 --> 01:05:57.120 align:middle line:90% are we choosing first? 01:05:57.120 --> 01:05:58.578 align:middle line:84% Well, we're not choosing the first. 01:05:58.578 --> 01:06:00.590 align:middle line:90% We're choosing them together. 01:06:00.590 --> 01:06:02.510 align:middle line:84% So we've gotten rid of that ambiguity 01:06:02.510 --> 01:06:04.410 align:middle line:90% by doing it in this way. 01:06:04.410 --> 01:06:06.140 align:middle line:84% And let's see how many ways there 01:06:06.140 --> 01:06:07.710 align:middle line:90% are to fill out this recipe. 01:06:07.710 --> 01:06:09.540 align:middle line:84% First, we have to choose a suit S. 01:06:09.540 --> 01:06:13.100 align:middle line:84% And there are four suits to choose from. 01:06:13.100 --> 01:06:18.870 align:middle line:84% Then we have to choose two cards with suit S. And there are 13. 01:06:18.870 --> 01:06:23.330 align:middle line:90% Choose two ways to do that. 01:06:23.330 --> 01:06:28.580 align:middle line:84% Now choose a card with suit T1, 13 ways, with suit T2, 13 ways, 01:06:28.580 --> 01:06:31.340 align:middle line:90% with T3, 13 ways. 01:06:31.340 --> 01:06:36.110 align:middle line:84% And so the answer is 4 times 13 choose 2 times 13 times 01:06:36.110 --> 01:06:38.720 align:middle line:90% 13 times 13. 01:06:38.720 --> 01:06:41.420 align:middle line:84% Which does look a little different from the answer 01:06:41.420 --> 01:06:42.180 align:middle line:90% we got up there. 01:06:42.180 --> 01:06:46.250 align:middle line:84% But if you check, they're actually the same answer. 01:06:46.250 --> 01:06:47.400 align:middle line:90% Does that make sense? 01:06:47.400 --> 01:06:48.710 align:middle line:90% Sound good? 01:06:48.710 --> 01:06:50.720 align:middle line:90% Wonderful. 01:06:50.720 --> 01:06:52.920 align:middle line:84% One last example I want to look at. 01:06:52.920 --> 01:07:01.945 align:middle line:84% How many hands-- how many hands have at least one pair? 01:07:01.945 --> 01:07:04.550 align:middle line:90% 01:07:04.550 --> 01:07:07.075 align:middle line:90% So pair means same rank. 01:07:07.075 --> 01:07:10.130 align:middle line:90% 01:07:10.130 --> 01:07:13.620 align:middle line:84% So how many hands have at least two cards of the same rank? 01:07:13.620 --> 01:07:17.250 align:middle line:84% So at least two queens, at least two sevens, something like that. 01:07:17.250 --> 01:07:19.620 align:middle line:84% Maybe you have three or four queens. 01:07:19.620 --> 01:07:23.010 align:middle line:84% Maybe you have two queens and two sevens, or three sevens. 01:07:23.010 --> 01:07:25.510 align:middle line:84% All of that is fine as long as at least two of your cards 01:07:25.510 --> 01:07:26.370 align:middle line:90% have the same rank. 01:07:26.370 --> 01:07:30.100 align:middle line:90% 01:07:30.100 --> 01:07:32.723 align:middle line:90% And here is my proposed recipe. 01:07:32.723 --> 01:07:37.420 align:middle line:90% 01:07:37.420 --> 01:07:40.530 align:middle line:90% Choose a rank for the pair. 01:07:40.530 --> 01:07:43.090 align:middle line:90% 01:07:43.090 --> 01:07:45.525 align:middle line:84% Let's call it choose rank R for the pair. 01:07:45.525 --> 01:07:48.550 align:middle line:90% 01:07:48.550 --> 01:07:59.505 align:middle line:84% Choose two cards with rank R. Choose three other cards. 01:07:59.505 --> 01:08:03.730 align:middle line:90% 01:08:03.730 --> 01:08:05.830 align:middle line:84% Here's a recipe to build up hands 01:08:05.830 --> 01:08:07.870 align:middle line:90% that have at least one pair. 01:08:07.870 --> 01:08:11.570 align:middle line:84% We know they have a pair because here's the pair in step two. 01:08:11.570 --> 01:08:15.590 align:middle line:84% Everything we build has at least one pair. 01:08:15.590 --> 01:08:18.609 align:middle line:84% And that's the first of our two important questions. 01:08:18.609 --> 01:08:22.149 align:middle line:84% Does everything we build have our property? 01:08:22.149 --> 01:08:24.760 align:middle line:90% The answer in this case is yes. 01:08:24.760 --> 01:08:27.760 align:middle line:84% Does everything we want to build come 01:08:27.760 --> 01:08:30.430 align:middle line:84% from our recipe in exactly one way 01:08:30.430 --> 01:08:32.400 align:middle line:84% is the other question we need to ask. 01:08:32.400 --> 01:08:44.439 align:middle line:90% 01:08:44.439 --> 01:08:53.260 align:middle line:84% Does every hand with at least one pair 01:08:53.260 --> 01:08:59.160 align:middle line:84% come from recipe in exactly one way? 01:08:59.160 --> 01:09:01.750 align:middle line:90% 01:09:01.750 --> 01:09:04.790 align:middle line:84% This is the other question we need to ask. 01:09:04.790 --> 01:09:07.569 align:middle line:84% Usually, this is the harder of the two questions. 01:09:07.569 --> 01:09:09.560 align:middle line:84% Because usually when trying to build our set, 01:09:09.560 --> 01:09:12.760 align:middle line:84% we're building it in order to have the property we care about. 01:09:12.760 --> 01:09:15.460 align:middle line:84% So that first of the two important questions 01:09:15.460 --> 01:09:18.520 align:middle line:90% is usually fine. 01:09:18.520 --> 01:09:21.609 align:middle line:90% Is this true? 01:09:21.609 --> 01:09:24.939 align:middle line:84% And let's look at an example here. 01:09:24.939 --> 01:09:33.220 align:middle line:84% So what if we want the hand ace of hearts, ace of spades, 01:09:33.220 --> 01:09:41.020 align:middle line:84% two of clubs, two of diamonds, three of hearts/ How can I get 01:09:41.020 --> 01:09:44.397 align:middle line:90% this hand with my recipe? 01:09:44.397 --> 01:09:45.480 align:middle line:90% Can someone give me a way? 01:09:45.480 --> 01:09:49.149 align:middle line:90% 01:09:49.149 --> 01:09:50.800 align:middle line:90% Yep. 01:09:50.800 --> 01:09:52.389 align:middle line:84% AUDIENCE: You can choose [INAUDIBLE]. 01:09:52.389 --> 01:09:57.545 align:middle line:90% 01:09:57.545 --> 01:09:59.170 align:middle line:84% ZACHARY ABEL: OK, let's choose our rank 01:09:59.170 --> 01:10:03.080 align:middle line:90% as A for our first step. 01:10:03.080 --> 01:10:05.475 align:middle line:84% Second step is two cards with rank ace. 01:10:05.475 --> 01:10:08.890 align:middle line:90% 01:10:08.890 --> 01:10:13.660 align:middle line:84% Ace of hearts and ace of spades for our second step. 01:10:13.660 --> 01:10:15.890 align:middle line:84% And for our third step, the other three cards. 01:10:15.890 --> 01:10:19.550 align:middle line:84% In this case, that's two of clubs, two of diamonds, 01:10:19.550 --> 01:10:20.700 align:middle line:90% three of hearts. 01:10:20.700 --> 01:10:22.820 align:middle line:84% This is a way to fill out our recipe in order 01:10:22.820 --> 01:10:24.170 align:middle line:90% to get that hand. 01:10:24.170 --> 01:10:25.540 align:middle line:90% Who can give me another way? 01:10:25.540 --> 01:10:28.670 align:middle line:90% 01:10:28.670 --> 01:10:29.862 align:middle line:90% Yep. 01:10:29.862 --> 01:10:30.950 align:middle line:90% AUDIENCE: [INAUDIBLE] 01:10:30.950 --> 01:10:31.950 align:middle line:90% ZACHARY ABEL: All right. 01:10:31.950 --> 01:10:33.003 align:middle line:90% Rank equals 2. 01:10:33.003 --> 01:10:37.141 align:middle line:90% 01:10:37.141 --> 01:10:39.740 align:middle line:90% AUDIENCE: [INAUDIBLE] 01:10:39.740 --> 01:10:41.610 align:middle line:90% ZACHARY ABEL: All right. 01:10:41.610 --> 01:10:44.960 align:middle line:84% Then the two cards of rank two we're going to pick 01:10:44.960 --> 01:10:48.380 align:middle line:84% are going to be two of clubs and two of diamonds. 01:10:48.380 --> 01:10:50.042 align:middle line:84% And then the last step of our recipe 01:10:50.042 --> 01:10:52.250 align:middle line:84% is choose the other three cards, which, in this case, 01:10:52.250 --> 01:10:57.290 align:middle line:84% are ace of hearts, ace of spades, and three of hearts. 01:10:57.290 --> 01:10:59.600 align:middle line:84% So we found two different ways of filling out 01:10:59.600 --> 01:11:02.330 align:middle line:90% our recipe to get this hand. 01:11:02.330 --> 01:11:05.960 align:middle line:84% So we've already broken our second question. 01:11:05.960 --> 01:11:09.680 align:middle line:90% This answer is no. 01:11:09.680 --> 01:11:11.670 align:middle line:84% But we've already seen a possible backup. 01:11:11.670 --> 01:11:15.300 align:middle line:84% If it's always two to one, then we're fine. 01:11:15.300 --> 01:11:18.810 align:middle line:84% Then we just divide by 2 and get our answer. 01:11:18.810 --> 01:11:21.690 align:middle line:84% But let's look at a different example. 01:11:21.690 --> 01:11:28.240 align:middle line:84% What if our hand is instead ace of hearts, ace of diamonds, 01:11:28.240 --> 01:11:32.803 align:middle line:84% ace of clubs, three of clubs, five of spades? 01:11:32.803 --> 01:11:35.490 align:middle line:90% 01:11:35.490 --> 01:11:37.990 align:middle line:84% A recipe really breaks down into two phases. 01:11:37.990 --> 01:11:42.300 align:middle line:84% First, pick a pair, and then pick the other three cards. 01:11:42.300 --> 01:11:44.350 align:middle line:84% Pick the rank for our pair, then the-- sorry. 01:11:44.350 --> 01:11:46.600 align:middle line:84% Pick a suit for the pair, then the ranks for the pair. 01:11:46.600 --> 01:11:48.810 align:middle line:84% Really, pick a pair and then pick the other three 01:11:48.810 --> 01:11:51.400 align:middle line:90% cards is what our recipe says. 01:11:51.400 --> 01:11:55.150 align:middle line:84% Well, there are lots of ways to do that for this hand. 01:11:55.150 --> 01:11:59.220 align:middle line:84% We could pick this pair for our first two steps of the recipe 01:11:59.220 --> 01:12:01.830 align:middle line:84% and then choose the other three cards. 01:12:01.830 --> 01:12:06.510 align:middle line:84% Or we could pick that pair and then choose the other three 01:12:06.510 --> 01:12:07.510 align:middle line:90% cards. 01:12:07.510 --> 01:12:10.890 align:middle line:84% Or we could pick that pair and then choose the other three 01:12:10.890 --> 01:12:12.630 align:middle line:90% cards. 01:12:12.630 --> 01:12:18.383 align:middle line:84% So this one had two ways, this hand has three ways. 01:12:18.383 --> 01:12:21.930 align:middle line:90% 01:12:21.930 --> 01:12:25.410 align:middle line:90% Here's another example. 01:12:25.410 --> 01:12:27.040 align:middle line:90% Ace of diamonds-- 01:12:27.040 --> 01:12:27.600 align:middle line:90% Oops. 01:12:27.600 --> 01:12:32.500 align:middle line:84% Ace of clubs, three of clubs, and then three of spades. 01:12:32.500 --> 01:12:34.470 align:middle line:84% Same example as before, but now these 01:12:34.470 --> 01:12:36.970 align:middle line:84% are both 3's instead of a 3 and 5. 01:12:36.970 --> 01:12:39.190 align:middle line:90% How many pairs are there now? 01:12:39.190 --> 01:12:42.550 align:middle line:90% 1, 2, 3, 4 different pairs. 01:12:42.550 --> 01:12:48.150 align:middle line:84% So there are four ways to build this with our recipe. 01:12:48.150 --> 01:12:49.720 align:middle line:90% Let's keep going. 01:12:49.720 --> 01:12:52.050 align:middle line:84% What if we have a four of a kind? 01:12:52.050 --> 01:12:56.653 align:middle line:84% We have all four aces, and then we have the queen of diamonds. 01:12:56.653 --> 01:12:59.190 align:middle line:90% 01:12:59.190 --> 01:13:03.060 align:middle line:84% So that was clubs, hearts, spades, diamonds. 01:13:03.060 --> 01:13:08.200 align:middle line:84% Now there are 1, 2, 3, 4, 5, 6 different pairs. 01:13:08.200 --> 01:13:11.820 align:middle line:84% So there are six different ways to construct 01:13:11.820 --> 01:13:13.163 align:middle line:90% this hand with our recipe. 01:13:13.163 --> 01:13:16.210 align:middle line:90% 01:13:16.210 --> 01:13:18.740 align:middle line:84% Did all of these examples make sense? 01:13:18.740 --> 01:13:19.480 align:middle line:90% Cool. 01:13:19.480 --> 01:13:23.440 align:middle line:84% And what we've learned here is that this recipe 01:13:23.440 --> 01:13:25.390 align:middle line:90% is utter garbage. 01:13:25.390 --> 01:13:27.560 align:middle line:90% This recipe does not help. 01:13:27.560 --> 01:13:30.340 align:middle line:84% It doesn't make a bijection because there are multiple ways 01:13:30.340 --> 01:13:31.730 align:middle line:90% to build this hand. 01:13:31.730 --> 01:13:33.680 align:middle line:84% It doesn't make a k to 1 mapping because there 01:13:33.680 --> 01:13:35.680 align:middle line:84% are different number of ways, depending on which 01:13:35.680 --> 01:13:37.190 align:middle line:90% hand we're trying to build. 01:13:37.190 --> 01:13:42.100 align:middle line:84% So we can't just count this and then divide by 2 or divide by 3. 01:13:42.100 --> 01:13:42.910 align:middle line:90% Question? 01:13:42.910 --> 01:13:45.850 align:middle line:84% AUDIENCE: Sorry, how did you get from four ways to six ways? 01:13:45.850 --> 01:13:46.690 align:middle line:90% ZACHARY ABEL: Yeah. 01:13:46.690 --> 01:13:53.290 align:middle line:84% So with this hand, steps one and two really say pick a pair. 01:13:53.290 --> 01:13:55.580 align:middle line:84% And then step three says pick the other three cards. 01:13:55.580 --> 01:13:58.330 align:middle line:84% Well, how many pairs can we pick in this hand? 01:13:58.330 --> 01:14:04.190 align:middle line:84% There are four choose two ways to pick two of the aces. 01:14:04.190 --> 01:14:07.720 align:middle line:90% 4 choose 2 is 6. 01:14:07.720 --> 01:14:08.990 align:middle line:90% Or we can just look at it. 01:14:08.990 --> 01:14:14.570 align:middle line:90% It's 1, 2, 3, 4, 5, 6. 01:14:14.570 --> 01:14:17.120 align:middle line:90% Those are all the pairs. 01:14:17.120 --> 01:14:17.620 align:middle line:90% Yeah. 01:14:17.620 --> 01:14:18.680 align:middle line:90% Good question. 01:14:18.680 --> 01:14:22.550 align:middle line:90% So this recipe does not help. 01:14:22.550 --> 01:14:25.092 align:middle line:84% I don't care how many ways there are to fill out this recipe. 01:14:25.092 --> 01:14:27.633 align:middle line:84% I don't care how easy it is to count with generalized product 01:14:27.633 --> 01:14:28.160 align:middle line:90% rule. 01:14:28.160 --> 01:14:30.400 align:middle line:84% The thing it's counting is not at all 01:14:30.400 --> 01:14:32.800 align:middle line:84% related to the problem we care about. 01:14:32.800 --> 01:14:35.120 align:middle line:84% So this is the thing you need to watch out for. 01:14:35.120 --> 01:14:39.442 align:middle line:84% And that's why these two questions are 01:14:39.442 --> 01:14:41.650 align:middle line:84% the things you should ask yourself about every recipe 01:14:41.650 --> 01:14:42.590 align:middle line:90% you see. 01:14:42.590 --> 01:14:46.030 align:middle line:84% Can you build everything you care about? 01:14:46.030 --> 01:14:50.290 align:middle line:84% And is everything you care about built in exactly one way? 01:14:50.290 --> 01:14:52.600 align:middle line:84% And it takes some practice, and some skill, 01:14:52.600 --> 01:14:55.600 align:middle line:84% and some patience to figure out the right recipe in order 01:14:55.600 --> 01:14:57.890 align:middle line:84% to satisfy both of these constraints 01:14:57.890 --> 01:15:02.020 align:middle line:84% and to work out any of the ambiguities 01:15:02.020 --> 01:15:05.710 align:middle line:84% that we see in an example like this. 01:15:05.710 --> 01:15:06.298 align:middle line:90% All right. 01:15:06.298 --> 01:15:07.340 align:middle line:90% That's everything I have. 01:15:07.340 --> 01:15:09.240 align:middle line:90% Thank you so much. 01:15:09.240 --> 01:15:14.000 align:middle line:90%