WEBVTT 00:00:00.000 --> 00:00:00.500 align:middle line:90% 00:00:00.500 --> 00:00:02.000 align:middle line:90% [SQUEAKING] 00:00:02.000 --> 00:00:04.500 align:middle line:90% [RUSTLING] 00:00:04.500 --> 00:00:05.500 align:middle line:90% [CLICKING] 00:00:05.500 --> 00:00:11.000 align:middle line:90% 00:00:11.000 --> 00:00:12.600 align:middle line:84% BRYNMOR CHAPMAN: Hello, everybody. 00:00:12.600 --> 00:00:14.360 align:middle line:90% Can people hear me in the back? 00:00:14.360 --> 00:00:15.120 align:middle line:90% Is it working? 00:00:15.120 --> 00:00:16.100 align:middle line:90% Cool. 00:00:16.100 --> 00:00:22.100 align:middle line:84% OK, so I believe that last time, we started counting, 00:00:22.100 --> 00:00:26.030 align:middle line:84% and you saw the sum rule in particular. 00:00:26.030 --> 00:00:28.710 align:middle line:84% So let's take a look at a quick example. 00:00:28.710 --> 00:00:31.910 align:middle line:84% So I've got a very standard deck of playing cards here, 00:00:31.910 --> 00:00:33.740 align:middle line:90% as you can see. 00:00:33.740 --> 00:00:36.316 align:middle line:84% How many kings and queens are there? 00:00:36.316 --> 00:00:37.310 align:middle line:90% AUDIENCE: Eight. 00:00:37.310 --> 00:00:38.268 align:middle line:90% BRYNMOR CHAPMAN: Eight. 00:00:38.268 --> 00:00:39.485 align:middle line:90% Why is that? 00:00:39.485 --> 00:00:41.360 align:middle line:84% AUDIENCE: There's four kings and four queens. 00:00:41.360 --> 00:00:43.040 align:middle line:84% BRYNMOR CHAPMAN: Yeah, so the answer 00:00:43.040 --> 00:00:46.290 align:middle line:84% was that there are four suits for each rank. 00:00:46.290 --> 00:00:49.380 align:middle line:84% We've got, therefore, four kings, four queens. 00:00:49.380 --> 00:00:51.390 align:middle line:84% Nothing is both a king and a queen. 00:00:51.390 --> 00:00:52.980 align:middle line:84% So you can just add them together. 00:00:52.980 --> 00:00:57.380 align:middle line:84% You've got eight cards that are either kings or queens. 00:00:57.380 --> 00:00:59.550 align:middle line:84% What if the sets are not disjoint? 00:00:59.550 --> 00:01:01.910 align:middle line:84% So what if I instead wanted to count 00:01:01.910 --> 00:01:05.019 align:middle line:84% the number of cards that are either hearts or face cards? 00:01:05.019 --> 00:01:08.000 align:middle line:90% 00:01:08.000 --> 00:01:10.790 align:middle line:90% So then what do I do? 00:01:10.790 --> 00:01:14.610 align:middle line:90% So I've got 12 face cards here. 00:01:14.610 --> 00:01:16.860 align:middle line:84% My hands are a little bit too small for playing cards, 00:01:16.860 --> 00:01:19.962 align:middle line:84% but hopefully, you can roughly see. 00:01:19.962 --> 00:01:21.235 align:middle line:90% How many hearts are there? 00:01:21.235 --> 00:01:24.950 align:middle line:90% 00:01:24.950 --> 00:01:31.430 align:middle line:84% Yeah, 13 hearts, so 13 hearts in one hand, 12 face cards 00:01:31.430 --> 00:01:33.513 align:middle line:90% in the other, so 25? 00:01:33.513 --> 00:01:37.460 align:middle line:90% 00:01:37.460 --> 00:01:38.180 align:middle line:90% No? 00:01:38.180 --> 00:01:38.965 align:middle line:90% What's gone wrong? 00:01:38.965 --> 00:01:43.747 align:middle line:90% 00:01:43.747 --> 00:01:45.330 align:middle line:84% Can anybody tell me what's gone wrong? 00:01:45.330 --> 00:01:46.160 align:middle line:90% Yeah. 00:01:46.160 --> 00:01:48.850 align:middle line:84% AUDIENCE: You've got some cards that are both hearts and face 00:01:48.850 --> 00:01:49.350 align:middle line:90% cards. 00:01:49.350 --> 00:01:51.725 align:middle line:84% BRYNMOR CHAPMAN: Yeah, I've got two kings of hearts here, 00:01:51.725 --> 00:01:53.660 align:middle line:90% so I'm cheating. 00:01:53.660 --> 00:01:56.960 align:middle line:90% So what do? 00:01:56.960 --> 00:01:59.250 align:middle line:84% AUDIENCE: Well, since you're double-counting 00:01:59.250 --> 00:02:02.020 align:middle line:84% the cards that are both hearts and faces, 00:02:02.020 --> 00:02:03.938 align:middle line:84% you can subtract off one of them. 00:02:03.938 --> 00:02:05.230 align:middle line:90% BRYNMOR CHAPMAN: Yeah, exactly. 00:02:05.230 --> 00:02:07.650 align:middle line:84% So the answer was, I've got three cards that 00:02:07.650 --> 00:02:09.699 align:middle line:90% are both hearts and face cards. 00:02:09.699 --> 00:02:12.400 align:middle line:84% So these are the three cards that I've counted twice. 00:02:12.400 --> 00:02:15.990 align:middle line:84% I've got two each of these three cards. 00:02:15.990 --> 00:02:20.612 align:middle line:84% So if I just discard one copy, then I'll be fine. 00:02:20.612 --> 00:02:24.090 align:middle line:84% So I subtract off everything that's in the intersection. 00:02:24.090 --> 00:02:27.910 align:middle line:84% So I had 25 before, subtract off those three, 00:02:27.910 --> 00:02:30.750 align:middle line:84% and I end up with 22 that are either hearts or-- 00:02:30.750 --> 00:02:32.110 align:middle line:90% well, these cards are too big. 00:02:32.110 --> 00:02:34.440 align:middle line:84% Maybe I should have actually brought standard cards. 00:02:34.440 --> 00:02:38.410 align:middle line:84% I've got 22 cards that are either hearts or face cards. 00:02:38.410 --> 00:02:42.090 align:middle line:90% Does that make sense to people? 00:02:42.090 --> 00:02:45.870 align:middle line:84% So this is what we call the inclusion-exclusion rule 00:02:45.870 --> 00:02:51.030 align:middle line:84% or the principle of inclusion and exclusion, 00:02:51.030 --> 00:02:55.590 align:middle line:84% so often abbreviated as the PIE for Principle 00:02:55.590 --> 00:02:58.450 align:middle line:90% of Inclusion-Exclusion. 00:02:58.450 --> 00:03:06.534 align:middle line:84% So if I have 2 sets A and B and I want to compute their union, 00:03:06.534 --> 00:03:10.340 align:middle line:84% the size of this union-- ew, well, this chalk is terrible. 00:03:10.340 --> 00:03:15.200 align:middle line:84% Sorry-- the size of their union is 00:03:15.200 --> 00:03:21.260 align:middle line:84% going to be the sizes of the individual sets minus the size 00:03:21.260 --> 00:03:22.500 align:middle line:90% of the intersection. 00:03:22.500 --> 00:03:26.825 align:middle line:90% 00:03:26.825 --> 00:03:29.472 align:middle line:84% Does that make sense to everybody? 00:03:29.472 --> 00:03:33.914 align:middle line:84% It may be a little bit easier to see if we draw a picture. 00:03:33.914 --> 00:03:40.770 align:middle line:84% So hopefully, people remember Venn diagrams from grade school. 00:03:40.770 --> 00:03:41.960 align:middle line:90% Yeah? 00:03:41.960 --> 00:03:42.750 align:middle line:90% Maybe? 00:03:42.750 --> 00:03:43.250 align:middle line:90% No? 00:03:43.250 --> 00:03:46.710 align:middle line:84% Well, this is a Venn diagram, a couple of overlapping circles. 00:03:46.710 --> 00:03:49.220 align:middle line:84% So maybe we've got hearts on one side, 00:03:49.220 --> 00:03:52.290 align:middle line:84% and we've got faces on the other. 00:03:52.290 --> 00:03:54.540 align:middle line:84% But then we've got some things that are in the middle. 00:03:54.540 --> 00:03:57.560 align:middle line:84% So this can be our set A. This can be our set B. 00:03:57.560 --> 00:04:01.840 align:middle line:84% And here, we've got king of hearts, queen of hearts, 00:04:01.840 --> 00:04:05.605 align:middle line:90% and jack of hearts. 00:04:05.605 --> 00:04:08.420 align:middle line:84% So if we want to compute the size of the union, 00:04:08.420 --> 00:04:11.770 align:middle line:84% we're going to take the things that are only on the heart side, 00:04:11.770 --> 00:04:13.520 align:middle line:84% the things that are only on the face side, 00:04:13.520 --> 00:04:15.562 align:middle line:84% and then the things that are in the intersection, 00:04:15.562 --> 00:04:17.382 align:middle line:90% we only want to count them once. 00:04:17.382 --> 00:04:19.515 align:middle line:84% But that's counted in both of A and B. 00:04:19.515 --> 00:04:21.640 align:middle line:84% So we're going to add together the sizes of A and B 00:04:21.640 --> 00:04:24.100 align:middle line:84% and then subtract out the middle Does that 00:04:24.100 --> 00:04:27.710 align:middle line:90% make sense to everybody? 00:04:27.710 --> 00:04:31.030 align:middle line:84% So what happens if we have three sets? 00:04:31.030 --> 00:04:31.725 align:middle line:90% Any ideas? 00:04:31.725 --> 00:04:41.230 align:middle line:90% 00:04:41.230 --> 00:04:43.090 align:middle line:84% Well, can we do something similar? 00:04:43.090 --> 00:04:45.975 align:middle line:84% So let's draw a three-way Venn diagram this time. 00:04:45.975 --> 00:04:49.490 align:middle line:90% 00:04:49.490 --> 00:04:54.180 align:middle line:84% So we've got A here, B here, C here. 00:04:54.180 --> 00:05:01.020 align:middle line:84% So is it true that A union B union C 00:05:01.020 --> 00:05:07.140 align:middle line:84% is equal to the sum of the individual sets 00:05:07.140 --> 00:05:09.430 align:middle line:84% minus the pairwise intersections? 00:05:09.430 --> 00:05:16.470 align:middle line:90% 00:05:16.470 --> 00:05:25.890 align:middle line:84% Oops, that should be a B. Is this true? 00:05:25.890 --> 00:05:27.740 align:middle line:90% Yeah. 00:05:27.740 --> 00:05:41.010 align:middle line:84% AUDIENCE: [INAUDIBLE] A and B, A and C, B and C [INAUDIBLE] 00:05:41.010 --> 00:05:41.680 align:middle line:90% overlap. 00:05:41.680 --> 00:05:43.780 align:middle line:90% BRYNMOR CHAPMAN: Yeah. 00:05:43.780 --> 00:05:47.460 align:middle line:84% So your classmate's observation was that these intersections 00:05:47.460 --> 00:05:50.284 align:middle line:90% themselves overlap. 00:05:50.284 --> 00:05:55.780 align:middle line:84% So if we think about how many times we've counted each item-- 00:05:55.780 --> 00:05:57.752 align:middle line:84% everything here, we've only counted once. 00:05:57.752 --> 00:05:59.460 align:middle line:84% Everything here, we've only counted once. 00:05:59.460 --> 00:06:01.200 align:middle line:84% Everything here, we've only counted once. 00:06:01.200 --> 00:06:03.710 align:middle line:90% That's fairly easy. 00:06:03.710 --> 00:06:09.112 align:middle line:84% What about the things in these kind of weirdly shaped segments? 00:06:09.112 --> 00:06:12.530 align:middle line:84% So those appear in two of the individual sets 00:06:12.530 --> 00:06:15.420 align:middle line:84% and in one of the pairwise intersections. 00:06:15.420 --> 00:06:18.990 align:middle line:84% So we've also counted these ones. 00:06:18.990 --> 00:06:21.662 align:middle line:84% But now we've got a problem in the middle. 00:06:21.662 --> 00:06:24.020 align:middle line:84% How many times have we counted this? 00:06:24.020 --> 00:06:27.590 align:middle line:84% Well, we've counted it once for A, once for B, once for C, 00:06:27.590 --> 00:06:29.300 align:middle line:84% and then we've subtracted it once 00:06:29.300 --> 00:06:33.690 align:middle line:84% each for A intersect B, A intersect C, and B intersect C. 00:06:33.690 --> 00:06:35.780 align:middle line:84% So we've counted this zero times in total. 00:06:35.780 --> 00:06:38.450 align:middle line:90% 00:06:38.450 --> 00:06:41.090 align:middle line:84% So what we actually want is to add back 00:06:41.090 --> 00:06:48.662 align:middle line:84% in the three-way intersection, and that gives us 00:06:48.662 --> 00:06:49.495 align:middle line:90% the correct formula. 00:06:49.495 --> 00:06:52.144 align:middle line:90% 00:06:52.144 --> 00:06:55.250 align:middle line:84% Does that make sense to everybody? 00:06:55.250 --> 00:06:57.720 align:middle line:90% Yeah? 00:06:57.720 --> 00:06:59.950 align:middle line:84% So let's take a look at an example then. 00:06:59.950 --> 00:07:06.500 align:middle line:90% 00:07:06.500 --> 00:07:17.740 align:middle line:84% How many numbers k in the range from 1 to n are coprime with n? 00:07:17.740 --> 00:07:32.420 align:middle line:90% 00:07:32.420 --> 00:07:35.078 align:middle line:90% How could we figure this out? 00:07:35.078 --> 00:07:37.370 align:middle line:84% And those of you who already know the totient function, 00:07:37.370 --> 00:07:38.245 align:middle line:90% don't invoke that. 00:07:38.245 --> 00:07:45.050 align:middle line:90% 00:07:45.050 --> 00:07:47.810 align:middle line:84% How could we use the principle of inclusion-exclusion 00:07:47.810 --> 00:07:48.900 align:middle line:90% to compute this? 00:07:48.900 --> 00:07:49.665 align:middle line:90% Any ideas? 00:07:49.665 --> 00:07:53.920 align:middle line:90% 00:07:53.920 --> 00:07:55.040 align:middle line:90% Tentative hands? 00:07:55.040 --> 00:07:57.340 align:middle line:90% Maybe? 00:07:57.340 --> 00:07:58.310 align:middle line:90% Yeah. 00:07:58.310 --> 00:08:03.130 align:middle line:84% AUDIENCE: I guess you can do the prime factorization of n 00:08:03.130 --> 00:08:07.030 align:middle line:84% and then check what numbers within-- 00:08:07.030 --> 00:08:11.740 align:middle line:84% well, I don't know the specifics but basically check 00:08:11.740 --> 00:08:17.200 align:middle line:84% which numbers are multiples of a specific factor 00:08:17.200 --> 00:08:20.590 align:middle line:84% and then count which are the ones that 00:08:20.590 --> 00:08:22.460 align:middle line:90% share factors between them. 00:08:22.460 --> 00:08:25.480 align:middle line:90% 00:08:25.480 --> 00:08:29.500 align:middle line:84% BRYNMOR CHAPMAN: So the idea was look at which numbers 00:08:29.500 --> 00:08:31.720 align:middle line:90% share a prime factor with n. 00:08:31.720 --> 00:08:34.450 align:middle line:84% So for simplicity, let's assume for now 00:08:34.450 --> 00:08:38.019 align:middle line:84% that n is the product of, say, three primes, p, q, and r. 00:08:38.019 --> 00:08:42.070 align:middle line:90% 00:08:42.070 --> 00:08:45.310 align:middle line:84% So then how could we figure out how many 00:08:45.310 --> 00:08:46.620 align:middle line:90% things share a factor of p? 00:08:46.620 --> 00:08:56.340 align:middle line:90% 00:08:56.340 --> 00:08:57.810 align:middle line:90% Yeah. 00:08:57.810 --> 00:09:00.373 align:middle line:84% AUDIENCE: The quotient when you divide n by p. 00:09:00.373 --> 00:09:01.290 align:middle line:90% BRYNMOR CHAPMAN: Yeah. 00:09:01.290 --> 00:09:06.850 align:middle line:84% So there are going to be n/p that share a factor of p. 00:09:06.850 --> 00:09:10.530 align:middle line:90% So let's say that-- 00:09:10.530 --> 00:09:16.680 align:middle line:84% let's call this set A sub p, the set of numbers-- 00:09:16.680 --> 00:09:20.430 align:middle line:84% the subset of these numbers that share a factor of p. 00:09:20.430 --> 00:09:22.620 align:middle line:84% So there are going to be n/p of those. 00:09:22.620 --> 00:09:26.580 align:middle line:84% What about the things that share a factor of q? 00:09:26.580 --> 00:09:28.680 align:middle line:90% Same thing. 00:09:28.680 --> 00:09:31.110 align:middle line:84% So similarly, we can define A sub q 00:09:31.110 --> 00:09:34.890 align:middle line:84% to be the set of numbers that share a factor of q. 00:09:34.890 --> 00:09:38.460 align:middle line:84% And then we've got this, and similarly, A sub r 00:09:38.460 --> 00:09:41.015 align:middle line:84% is going to be the set of things that share a factor of r. 00:09:41.015 --> 00:09:49.610 align:middle line:90% 00:09:49.610 --> 00:09:51.425 align:middle line:84% Now, what exactly are we trying to compute? 00:09:51.425 --> 00:09:58.020 align:middle line:90% 00:09:58.020 --> 00:10:00.660 align:middle line:84% What we really want is the complement 00:10:00.660 --> 00:10:03.330 align:middle line:84% of the union of these three sets. 00:10:03.330 --> 00:10:16.200 align:middle line:84% So we're trying to compute the size of n minus Ap union Aq 00:10:16.200 --> 00:10:17.970 align:middle line:90% union Ar. 00:10:17.970 --> 00:10:31.390 align:middle line:90% 00:10:31.390 --> 00:10:34.050 align:middle line:84% So now we can use inclusion-exclusion. 00:10:34.050 --> 00:10:37.150 align:middle line:84% We're trying to compute the union of this set. 00:10:37.150 --> 00:10:41.550 align:middle line:84% That's going to be the sum of the sizes of the individual sets 00:10:41.550 --> 00:10:43.320 align:middle line:90% minus the-- 00:10:43.320 --> 00:10:47.710 align:middle line:84% oh, I did this the wrong way around, didn't I? 00:10:47.710 --> 00:10:48.440 align:middle line:90% Oh, no, I didn't. 00:10:48.440 --> 00:10:50.320 align:middle line:90% That was the right one. 00:10:50.320 --> 00:10:53.230 align:middle line:84% So the sum of the individual sizes 00:10:53.230 --> 00:10:57.250 align:middle line:84% minus the sum of the pairwise intersections plus the sum 00:10:57.250 --> 00:10:59.410 align:middle line:84% of the three-way intersection-- so 00:10:59.410 --> 00:11:01.550 align:middle line:84% what are the sizes of these intersections? 00:11:01.550 --> 00:11:20.680 align:middle line:90% 00:11:20.680 --> 00:11:22.905 align:middle line:84% What's the size of the intersection of Ap and Aq? 00:11:22.905 --> 00:11:29.620 align:middle line:90% 00:11:29.620 --> 00:11:30.400 align:middle line:90% Yeah. 00:11:30.400 --> 00:11:31.480 align:middle line:90% AUDIENCE: n over pq. 00:11:31.480 --> 00:11:32.800 align:middle line:90% BRYNMOR CHAPMAN: Yeah. 00:11:32.800 --> 00:11:34.690 align:middle line:90% So the answer was n over pq. 00:11:34.690 --> 00:11:37.640 align:middle line:90% 00:11:37.640 --> 00:11:39.830 align:middle line:90% So p and q are distinct primes. 00:11:39.830 --> 00:11:43.990 align:middle line:84% If something shares both the factor of p and a factor of q, 00:11:43.990 --> 00:11:46.840 align:middle line:90% it has to be a multiple of pq. 00:11:46.840 --> 00:11:49.650 align:middle line:90% So there are n over pq of those. 00:11:49.650 --> 00:11:58.950 align:middle line:84% Similarly, we have n over pr and n over qr. 00:11:58.950 --> 00:12:01.320 align:middle line:84% What about the three-way intersection? 00:12:01.320 --> 00:12:06.010 align:middle line:84% How many things share a factor of p, q, and r? 00:12:06.010 --> 00:12:06.510 align:middle line:90% Yeah. 00:12:06.510 --> 00:12:07.510 align:middle line:90% AUDIENCE: That's just n. 00:12:07.510 --> 00:12:10.590 align:middle line:84% BRYNMOR CHAPMAN: Yeah, it's just n itself, so minus 1. 00:12:10.590 --> 00:12:19.470 align:middle line:90% 00:12:19.470 --> 00:12:23.010 align:middle line:84% And if you do some messy algebra, 00:12:23.010 --> 00:12:32.160 align:middle line:84% this turns out to p minus 1, q minus 1, r minus 1, 00:12:32.160 --> 00:12:41.824 align:middle line:84% which is also equal to n times 1 minus 1/p, 1 minus 1/q, 00:12:41.824 --> 00:12:43.130 align:middle line:90% 1 minus 1/r. 00:12:43.130 --> 00:12:48.957 align:middle line:90% 00:12:48.957 --> 00:12:50.540 align:middle line:84% So hopefully, it's not too hard to see 00:12:50.540 --> 00:12:55.310 align:middle line:84% how this could generalize to get the more general form of Euler's 00:12:55.310 --> 00:12:57.860 align:middle line:90% totient. 00:12:57.860 --> 00:13:02.390 align:middle line:84% Does this application make sense to people? 00:13:02.390 --> 00:13:03.240 align:middle line:90% Any questions? 00:13:03.240 --> 00:13:04.224 align:middle line:90% Yeah. 00:13:04.224 --> 00:13:06.132 align:middle line:90% AUDIENCE: [INAUDIBLE]? 00:13:06.132 --> 00:13:08.090 align:middle line:84% BRYNMOR CHAPMAN: Oh, that's just the assumption 00:13:08.090 --> 00:13:11.440 align:middle line:84% that we made for the sake of example. 00:13:11.440 --> 00:13:25.510 align:middle line:90% 00:13:25.510 --> 00:13:30.300 align:middle line:84% So that's inclusion-exclusion with two or three sets, 00:13:30.300 --> 00:13:33.510 align:middle line:84% but more generally, you can continue the pattern. 00:13:33.510 --> 00:13:42.220 align:middle line:84% So if you have, say, n sets A1, A2, all the way up to A sub n 00:13:42.220 --> 00:13:56.070 align:middle line:84% and you want to figure out the union of all of these, 00:13:56.070 --> 00:14:00.460 align:middle line:84% so as we did before, we're going to add up the individual sizes 00:14:00.460 --> 00:14:02.910 align:middle line:90% of the individual sets. 00:14:02.910 --> 00:14:09.430 align:middle line:90% 00:14:09.430 --> 00:14:12.106 align:middle line:84% Then we're going to subtract off the pairwise intersections. 00:14:12.106 --> 00:14:22.081 align:middle line:90% 00:14:22.081 --> 00:14:36.420 align:middle line:84% Then we're going to add in the three-way intersections, 00:14:36.420 --> 00:14:40.210 align:middle line:84% and we're just going to keep on going until eventually we 00:14:40.210 --> 00:14:45.080 align:middle line:84% get to plus or minus the size of the n-way intersection, 00:14:45.080 --> 00:14:50.640 align:middle line:84% so A1 cap A2 cap all the way to A sub n. 00:14:50.640 --> 00:14:54.490 align:middle line:90% 00:14:54.490 --> 00:15:00.100 align:middle line:84% And the sign on that last term will depend on the parity of n. 00:15:00.100 --> 00:15:01.425 align:middle line:90% Does that make sense to people? 00:15:01.425 --> 00:15:11.290 align:middle line:90% 00:15:11.290 --> 00:15:16.630 align:middle line:84% So perhaps a more formal way of expressing this 00:15:16.630 --> 00:15:17.900 align:middle line:90% is the following. 00:15:17.900 --> 00:15:21.220 align:middle line:90% 00:15:21.220 --> 00:15:27.280 align:middle line:84% So suppose we have some set u, which we're going to say 00:15:27.280 --> 00:15:32.710 align:middle line:90% is a union of n sets. 00:15:32.710 --> 00:15:36.160 align:middle line:90% 00:15:36.160 --> 00:15:40.534 align:middle line:84% So this is going to be our universe of discourse. 00:15:40.534 --> 00:15:43.550 align:middle line:84% And so this means that by convention, 00:15:43.550 --> 00:15:48.565 align:middle line:84% if we intersect no sets, this is just going to be u. 00:15:48.565 --> 00:15:52.040 align:middle line:90% 00:15:52.040 --> 00:15:55.520 align:middle line:84% So just like if we union no sets, we start with nothing 00:15:55.520 --> 00:15:57.990 align:middle line:84% and then start unioning in more things, 00:15:57.990 --> 00:15:59.527 align:middle line:84% and we should get a union of sets. 00:15:59.527 --> 00:16:01.860 align:middle line:84% If we start with everything and then start intersecting, 00:16:01.860 --> 00:16:03.300 align:middle line:84% we should get an intersection of sets. 00:16:03.300 --> 00:16:05.190 align:middle line:84% So that's where this convention comes from. 00:16:05.190 --> 00:16:06.562 align:middle line:90% Yeah. 00:16:06.562 --> 00:16:09.830 align:middle line:90% AUDIENCE: [INAUDIBLE] 00:16:09.830 --> 00:16:12.270 align:middle line:84% BRYNMOR CHAPMAN: Oh, this notation? 00:16:12.270 --> 00:16:13.260 align:middle line:90% This notation? 00:16:13.260 --> 00:16:13.790 align:middle line:90% Both? 00:16:13.790 --> 00:16:14.570 align:middle line:90% AUDIENCE: Both [INAUDIBLE]. 00:16:14.570 --> 00:16:15.570 align:middle line:90% BRYNMOR CHAPMAN: Oh, OK. 00:16:15.570 --> 00:16:18.870 align:middle line:84% So this is kind of like the big sigma. 00:16:18.870 --> 00:16:23.840 align:middle line:84% So just how this represents like an n-way addition, 00:16:23.840 --> 00:16:27.410 align:middle line:84% this represents an n-way union, and this represents 00:16:27.410 --> 00:16:29.090 align:middle line:90% an n-way intersection. 00:16:29.090 --> 00:16:31.700 align:middle line:90% 00:16:31.700 --> 00:16:36.740 align:middle line:84% So we've got-- this is saying we've got a union of n sets, 00:16:36.740 --> 00:16:43.980 align:middle line:84% like A1 union A2 union A3 union An. 00:16:43.980 --> 00:16:48.698 align:middle line:90% 00:16:48.698 --> 00:16:52.180 align:middle line:84% And this is saying we've got an intersection of n sets, 00:16:52.180 --> 00:16:53.920 align:middle line:90% but n is zero. 00:16:53.920 --> 00:16:57.420 align:middle line:84% So by convention, we just take that to be the entire universe. 00:16:57.420 --> 00:16:59.918 align:middle line:90% AUDIENCE: [INAUDIBLE] 00:16:59.918 --> 00:17:00.876 align:middle line:90% BRYNMOR CHAPMAN: Sorry? 00:17:00.876 --> 00:17:03.680 align:middle line:84% AUDIENCE: What was the [INAUDIBLE] intersection 00:17:03.680 --> 00:17:04.855 align:middle line:90% [INAUDIBLE]? 00:17:04.855 --> 00:17:06.230 align:middle line:84% BRYNMOR CHAPMAN: The question is, 00:17:06.230 --> 00:17:08.480 align:middle line:84% why are we setting both the union and the intersection 00:17:08.480 --> 00:17:10.150 align:middle line:90% to be u? 00:17:10.150 --> 00:17:12.020 align:middle line:84% This is just how we're defining u. 00:17:12.020 --> 00:17:14.119 align:middle line:90% This is a convention. 00:17:14.119 --> 00:17:19.990 align:middle line:84% So if you intersect no sets, you should get everything, 00:17:19.990 --> 00:17:22.819 align:middle line:84% just like if you multiply no numbers, you should get 1. 00:17:22.819 --> 00:17:25.760 align:middle line:90% It's kind of similar to that. 00:17:25.760 --> 00:17:29.150 align:middle line:84% You start with everything, and then you intersect one set. 00:17:29.150 --> 00:17:33.530 align:middle line:84% You should get the set that you intersected in. 00:17:33.530 --> 00:17:35.260 align:middle line:90% So this is just a convention. 00:17:35.260 --> 00:17:39.220 align:middle line:84% If you prefer, you can just write in u itself. 00:17:39.220 --> 00:17:41.460 align:middle line:84% And what I'm about to write that might make it easier 00:17:41.460 --> 00:17:42.310 align:middle line:90% to understand. 00:17:42.310 --> 00:17:43.145 align:middle line:90% Sorry about that. 00:17:43.145 --> 00:17:51.513 align:middle line:90% 00:17:51.513 --> 00:17:59.640 align:middle line:84% So the theorem is that if you look 00:17:59.640 --> 00:18:04.560 align:middle line:84% at all subsets of the numbers up to n 00:18:04.560 --> 00:18:19.611 align:middle line:84% and you take the i-way intersection of these Ai's 00:18:19.611 --> 00:18:24.907 align:middle line:84% and then multiply it by minus 1 to the size of i, 00:18:24.907 --> 00:18:25.865 align:middle line:90% this should give you 0. 00:18:25.865 --> 00:18:32.040 align:middle line:90% 00:18:32.040 --> 00:18:36.020 align:middle line:84% So now, if you want to think about this in the case 00:18:36.020 --> 00:18:40.030 align:middle line:84% of the two-way inclusion-exclusion here-- 00:18:40.030 --> 00:18:42.750 align:middle line:90% 00:18:42.750 --> 00:18:46.160 align:middle line:90% so this is saying, 0 equals-- 00:18:46.160 --> 00:18:50.450 align:middle line:84% we move the union over to the right-hand side, 00:18:50.450 --> 00:18:52.100 align:middle line:84% or I guess we're moving everything over 00:18:52.100 --> 00:18:56.060 align:middle line:84% to the left-hand side based on what I've written. 00:18:56.060 --> 00:18:57.200 align:middle line:90% Yeah. 00:18:57.200 --> 00:19:02.765 align:middle line:90% AUDIENCE: [INAUDIBLE] 00:19:02.765 --> 00:19:04.640 align:middle line:84% BRYNMOR CHAPMAN: So this formula is basically 00:19:04.640 --> 00:19:05.940 align:middle line:90% what we were doing here. 00:19:05.940 --> 00:19:10.190 align:middle line:84% We're taking i-way intersections, 00:19:10.190 --> 00:19:15.360 align:middle line:84% and we're flipping the sign on successive sizes of i. 00:19:15.360 --> 00:19:17.930 align:middle line:84% And we're just kind of moving this union over 00:19:17.930 --> 00:19:19.945 align:middle line:84% to the other side, so we get zero. 00:19:19.945 --> 00:19:33.190 align:middle line:90% 00:19:33.190 --> 00:19:35.820 align:middle line:84% So if that formula is too much, you 00:19:35.820 --> 00:19:37.370 align:middle line:84% can think about the previous formula. 00:19:37.370 --> 00:19:38.600 align:middle line:90% But yeah. 00:19:38.600 --> 00:19:43.600 align:middle line:84% AUDIENCE: [INAUDIBLE] i so big I would [INAUDIBLE]. 00:19:43.600 --> 00:19:47.080 align:middle line:90% 00:19:47.080 --> 00:19:48.900 align:middle line:84% BRYNMOR CHAPMAN: So the question was, 00:19:48.900 --> 00:19:50.900 align:middle line:84% what is the difference between big and little i? 00:19:50.900 --> 00:19:53.750 align:middle line:90% So big I is a set of indices. 00:19:53.750 --> 00:19:56.370 align:middle line:84% Little i is an individual index in that set. 00:19:56.370 --> 00:19:59.956 align:middle line:90% 00:19:59.956 --> 00:20:08.800 align:middle line:84% So we're summing over all subsets of the numbers up to n, 00:20:08.800 --> 00:20:11.140 align:middle line:84% and then for each of those subsets, 00:20:11.140 --> 00:20:16.300 align:middle line:84% we're taking the intersection of the relevant Ai's. 00:20:16.300 --> 00:20:21.070 align:middle line:84% So over here, we've got A is just-- it corresponds 00:20:21.070 --> 00:20:23.782 align:middle line:90% to the subset just 1. 00:20:23.782 --> 00:20:26.680 align:middle line:90% B corresponds to the subset 2. 00:20:26.680 --> 00:20:30.910 align:middle line:84% A intersect B corresponds to the subset 1 comma 2, 00:20:30.910 --> 00:20:34.690 align:middle line:84% and A union B corresponds to the empty subset. 00:20:34.690 --> 00:20:38.650 align:middle line:84% So these are all of the subsets of the integers 1 and 2. 00:20:38.650 --> 00:20:40.210 align:middle line:90% Does that make sense? 00:20:40.210 --> 00:20:41.680 align:middle line:90% Question? 00:20:41.680 --> 00:20:45.290 align:middle line:84% AUDIENCE: Can you use I [INAUDIBLE] for negative 1? 00:20:45.290 --> 00:20:47.320 align:middle line:90% Why is it absolute value? 00:20:47.320 --> 00:20:50.530 align:middle line:84% BRYNMOR CHAPMAN: Size of I, because I is a set. 00:20:50.530 --> 00:20:53.570 align:middle line:84% Yeah, so the question is, why do we have absolute value here? 00:20:53.570 --> 00:20:55.250 align:middle line:90% That's the cardinality. 00:20:55.250 --> 00:21:11.080 align:middle line:90% 00:21:11.080 --> 00:21:12.570 align:middle line:90% So how could we prove this? 00:21:12.570 --> 00:21:17.680 align:middle line:90% 00:21:17.680 --> 00:21:20.600 align:middle line:90% Well, let's look at some x. 00:21:20.600 --> 00:21:25.670 align:middle line:84% So say that we've got some element x of our universe. 00:21:25.670 --> 00:21:28.410 align:middle line:90% 00:21:28.410 --> 00:21:34.675 align:middle line:84% How many times do we count x on this left-hand side? 00:21:34.675 --> 00:21:36.425 align:middle line:84% How many of these terms does it appear in? 00:21:36.425 --> 00:21:44.220 align:middle line:90% 00:21:44.220 --> 00:21:48.030 align:middle line:84% Well, it's going to be in all of the terms where x is in all 00:21:48.030 --> 00:21:50.280 align:middle line:90% of these Ai's. 00:21:50.280 --> 00:22:14.070 align:middle line:84% So if we say I sub x is the set of indices with x in Ai, 00:22:14.070 --> 00:22:18.270 align:middle line:84% it's going to be in the I sub x set. 00:22:18.270 --> 00:22:28.880 align:middle line:84% It's also going to be in every subset of I sub X. 00:22:28.880 --> 00:22:29.910 align:middle line:90% Does that make sense? 00:22:29.910 --> 00:22:31.970 align:middle line:90% Do people see why? 00:22:31.970 --> 00:22:38.900 align:middle line:84% If x is in sets 1 and 2, it's going to be in the intersection 00:22:38.900 --> 00:22:40.800 align:middle line:90% where we've got I is 1, 2. 00:22:40.800 --> 00:22:43.460 align:middle line:84% It's also going to be in the intersection where I is 1 00:22:43.460 --> 00:22:45.575 align:middle line:84% or where I is 2 or where I is empty. 00:22:45.575 --> 00:22:52.383 align:middle line:90% 00:22:52.383 --> 00:22:54.800 align:middle line:84% But now, how much does it contribute to the left-hand side 00:22:54.800 --> 00:22:55.550 align:middle line:90% for each of these? 00:22:55.550 --> 00:22:59.540 align:middle line:90% 00:22:59.540 --> 00:23:03.290 align:middle line:84% Well, how many subsets of I sub x have even size, 00:23:03.290 --> 00:23:04.420 align:middle line:90% and how many have odd size? 00:23:04.420 --> 00:23:10.005 align:middle line:90% 00:23:10.005 --> 00:23:10.505 align:middle line:90% Yeah. 00:23:10.505 --> 00:23:11.130 align:middle line:90% AUDIENCE: Half. 00:23:11.130 --> 00:23:14.120 align:middle line:84% BRYNMOR CHAPMAN: Yeah, it's half each. 00:23:14.120 --> 00:23:17.930 align:middle line:84% So for the ones of odd size, we're contributing minus 1 00:23:17.930 --> 00:23:19.170 align:middle line:90% to the left-hand side. 00:23:19.170 --> 00:23:21.620 align:middle line:84% For the ones of even size, we're contributing positive 1 00:23:21.620 --> 00:23:25.430 align:middle line:84% to the right-hand side-- or to the left-hand side. 00:23:25.430 --> 00:23:27.555 align:middle line:84% So because they're half and half, those all cancel. 00:23:27.555 --> 00:23:33.052 align:middle line:90% 00:23:33.052 --> 00:23:46.810 align:middle line:84% So x contributes plus 1 to the left-hand side 00:23:46.810 --> 00:23:52.563 align:middle line:90% for each even-size subset. 00:23:52.563 --> 00:23:56.170 align:middle line:90% 00:23:56.170 --> 00:24:04.360 align:middle line:84% I sub x, and it contributes minus 1 00:24:04.360 --> 00:24:11.400 align:middle line:84% to the left-hand side for each odd-size subset. 00:24:11.400 --> 00:24:16.495 align:middle line:90% 00:24:16.495 --> 00:24:19.635 align:middle line:84% So when we add all of those together, we get 0. 00:24:19.635 --> 00:24:25.260 align:middle line:90% 00:24:25.260 --> 00:24:27.840 align:middle line:84% So every element of our universe contributes 0 00:24:27.840 --> 00:24:29.780 align:middle line:84% to the left-hand side, so we get 0. 00:24:29.780 --> 00:24:37.766 align:middle line:90% 00:24:37.766 --> 00:24:41.460 align:middle line:90% Does that make sense to people? 00:24:41.460 --> 00:24:42.930 align:middle line:90% Yeah. 00:24:42.930 --> 00:24:45.180 align:middle line:84% AUDIENCE: Both the left-hand side? 00:24:45.180 --> 00:24:46.530 align:middle line:90% BRYNMOR CHAPMAN: Yes. 00:24:46.530 --> 00:24:48.750 align:middle line:90% We're looking at the-- 00:24:48.750 --> 00:24:50.340 align:middle line:84% sorry, question was, are those both 00:24:50.340 --> 00:24:51.760 align:middle line:90% supposed to be left-hand side? 00:24:51.760 --> 00:24:55.950 align:middle line:84% We're looking at what x contributes to this sum. 00:24:55.950 --> 00:24:58.480 align:middle line:84% How many of these terms does it contribute positively to? 00:24:58.480 --> 00:25:00.780 align:middle line:84% How many does it contribute negatively to? 00:25:00.780 --> 00:25:05.100 align:middle line:84% So whenever the size of I is even, 00:25:05.100 --> 00:25:07.140 align:middle line:84% it's going to contribute positively if it's 00:25:07.140 --> 00:25:08.890 align:middle line:90% contained in that intersection. 00:25:08.890 --> 00:25:11.610 align:middle line:84% Whenever the size of I is odd, it will contribute negatively 00:25:11.610 --> 00:25:13.000 align:middle line:90% when it's in the intersection. 00:25:13.000 --> 00:25:15.690 align:middle line:90% 00:25:15.690 --> 00:25:16.190 align:middle line:90% Yeah. 00:25:16.190 --> 00:25:19.530 align:middle line:90% 00:25:19.530 --> 00:25:22.880 align:middle line:84% AUDIENCE: How did you get the empty set? 00:25:22.880 --> 00:25:28.372 align:middle line:84% How do we construct the big I so that [INAUDIBLE]? 00:25:28.372 --> 00:25:29.330 align:middle line:90% BRYNMOR CHAPMAN: Sorry. 00:25:29.330 --> 00:25:31.520 align:middle line:84% The question was, how do we construct a big I 00:25:31.520 --> 00:25:33.385 align:middle line:84% so that the intersection is the empty set? 00:25:33.385 --> 00:25:35.343 align:middle line:84% AUDIENCE: Yeah, I may have understood it wrong, 00:25:35.343 --> 00:25:39.140 align:middle line:84% but [INAUDIBLE] intersection big U 00:25:39.140 --> 00:25:41.666 align:middle line:84% and then subtract it off to get [INAUDIBLE]. 00:25:41.666 --> 00:25:44.390 align:middle line:90% 00:25:44.390 --> 00:25:47.300 align:middle line:84% BRYNMOR CHAPMAN: Let me see if I'm understanding. 00:25:47.300 --> 00:25:54.860 align:middle line:84% You are trying to find an I such that this is empty or such 00:25:54.860 --> 00:25:58.122 align:middle line:90% that this is empty. 00:25:58.122 --> 00:25:59.030 align:middle line:90% AUDIENCE: Second one. 00:25:59.030 --> 00:26:01.032 align:middle line:90% BRYNMOR CHAPMAN: This one. 00:26:01.032 --> 00:26:03.110 align:middle line:90% So we don't care. 00:26:03.110 --> 00:26:06.260 align:middle line:84% We are looking at the sets that contain x, 00:26:06.260 --> 00:26:07.520 align:middle line:90% so they should never be empty. 00:26:07.520 --> 00:26:13.910 align:middle line:90% 00:26:13.910 --> 00:26:17.900 align:middle line:84% We're looking at what we had here. 00:26:17.900 --> 00:26:19.600 align:middle line:84% How many times have we counted x? 00:26:19.600 --> 00:26:25.780 align:middle line:90% 00:26:25.780 --> 00:26:30.130 align:middle line:84% So if this is still confusing, please review the lecture notes. 00:26:30.130 --> 00:26:35.930 align:middle line:84% Or it's not important that the proof is not important. 00:26:35.930 --> 00:26:38.770 align:middle line:84% The most important thing is knowing how to use it. 00:26:38.770 --> 00:26:44.210 align:middle line:84% So you want to add together the single sets, 00:26:44.210 --> 00:26:46.400 align:middle line:84% subtract off the pairwise intersections, 00:26:46.400 --> 00:26:50.680 align:middle line:84% add together the three-way intersections, cetera. 00:26:50.680 --> 00:26:54.970 align:middle line:84% So let's move on to the second topic for today, 00:26:54.970 --> 00:26:56.300 align:middle line:90% pidgeyhole principle. 00:26:56.300 --> 00:27:05.224 align:middle line:90% 00:27:05.224 --> 00:27:07.625 align:middle line:84% Has anybody heard of this before? 00:27:07.625 --> 00:27:08.125 align:middle line:90% Yeah. 00:27:08.125 --> 00:27:11.080 align:middle line:90% 00:27:11.080 --> 00:27:13.630 align:middle line:84% So the pidgeyhole principle-- it gets its name 00:27:13.630 --> 00:27:15.440 align:middle line:90% from medieval times. 00:27:15.440 --> 00:27:19.110 align:middle line:84% You have a bunch of pidgeyholes that are 00:27:19.110 --> 00:27:21.120 align:middle line:90% supposed to contain pidgeys. 00:27:21.120 --> 00:27:23.500 align:middle line:84% But what if you have more pidgeys than pidgeyholes? 00:27:23.500 --> 00:27:25.380 align:middle line:90% What happens? 00:27:25.380 --> 00:27:26.950 align:middle line:90% Well, two of them have to share. 00:27:26.950 --> 00:27:29.050 align:middle line:90% Two of them get squished. 00:27:29.050 --> 00:27:32.850 align:middle line:84% So that's basically what the pidgeyhole principle is saying. 00:27:32.850 --> 00:27:42.300 align:middle line:84% If we have two sets A and B where the size of A is greater 00:27:42.300 --> 00:27:53.920 align:middle line:84% than the size of B and f from A to B is total, 00:27:53.920 --> 00:27:55.466 align:middle line:90% then f cannot be injective. 00:27:55.466 --> 00:28:13.974 align:middle line:90% 00:28:13.974 --> 00:28:20.790 align:middle line:84% So if is A set of pigeons, B is a set of pigeonholes, 00:28:20.790 --> 00:28:27.950 align:middle line:84% and f is saying, pigeon A lives in pigeonhole B, 00:28:27.950 --> 00:28:29.570 align:middle line:90% this should be a total function. 00:28:29.570 --> 00:28:31.620 align:middle line:90% Every pidgey should have a home. 00:28:31.620 --> 00:28:33.980 align:middle line:84% We don't want to have homeless pidgeys. 00:28:33.980 --> 00:28:37.490 align:middle line:84% So it's a total function, but that means 00:28:37.490 --> 00:28:39.110 align:middle line:84% that it can't be injective, meaning 00:28:39.110 --> 00:28:41.850 align:middle line:84% that there have to be two pigeons that share a pigeonhole. 00:28:41.850 --> 00:28:47.108 align:middle line:90% 00:28:47.108 --> 00:28:49.223 align:middle line:84% So let's look at a couple of examples. 00:28:49.223 --> 00:28:55.970 align:middle line:90% 00:28:55.970 --> 00:28:58.850 align:middle line:84% Do we still have 26 students in the room? 00:28:58.850 --> 00:28:59.510 align:middle line:90% Nice. 00:28:59.510 --> 00:29:01.540 align:middle line:84% Not too many students have dropped. 00:29:01.540 --> 00:29:04.650 align:middle line:84% So there are more than 26 students in the room, 00:29:04.650 --> 00:29:11.690 align:middle line:84% so at least two of you share the first letter of your first name. 00:29:11.690 --> 00:29:18.920 align:middle line:84% So we've got, well, at least 27 students 00:29:18.920 --> 00:29:21.055 align:middle line:90% and 26 letters of the alphabet. 00:29:21.055 --> 00:29:25.340 align:middle line:90% 00:29:25.340 --> 00:29:29.750 align:middle line:84% So if we map every student to the first letter 00:29:29.750 --> 00:29:33.150 align:middle line:84% of their first name, that's a total function, 00:29:33.150 --> 00:29:34.590 align:middle line:90% so it can't be injective. 00:29:34.590 --> 00:29:50.180 align:middle line:90% 00:29:50.180 --> 00:29:51.520 align:middle line:90% So two students must share. 00:29:51.520 --> 00:30:00.080 align:middle line:90% 00:30:00.080 --> 00:30:01.885 align:middle line:90% Does that example make sense? 00:30:01.885 --> 00:30:08.120 align:middle line:90% 00:30:08.120 --> 00:30:10.870 align:middle line:84% What if I've got n differently colored pairs of socks? 00:30:10.870 --> 00:30:14.402 align:middle line:90% 00:30:14.402 --> 00:30:16.950 align:middle line:90% Don't worry about it. 00:30:16.950 --> 00:30:18.480 align:middle line:84% How many socks do I need to pull out 00:30:18.480 --> 00:30:20.813 align:middle line:84% of my drawer to ensure that I get two of the same color? 00:30:20.813 --> 00:30:23.430 align:middle line:90% 00:30:23.430 --> 00:30:24.330 align:middle line:90% Sorry? 00:30:24.330 --> 00:30:25.170 align:middle line:90% AUDIENCE: Plus 1. 00:30:25.170 --> 00:30:27.430 align:middle line:84% BRYNMOR CHAPMAN: Yeah, the answer was n plus 1. 00:30:27.430 --> 00:30:31.260 align:middle line:90% So in my case, two. 00:30:31.260 --> 00:30:34.625 align:middle line:84% So if I've got n socks-- or, sorry, n colors of socks-- 00:30:34.625 --> 00:30:42.510 align:middle line:90% 00:30:42.510 --> 00:30:47.525 align:middle line:84% I need n plus 1 to guarantee a pair. 00:30:47.525 --> 00:30:56.820 align:middle line:90% 00:30:56.820 --> 00:31:01.398 align:middle line:84% So I've been a little bit hand-wavy with this one. 00:31:01.398 --> 00:31:02.940 align:middle line:84% When you're trying to prove something 00:31:02.940 --> 00:31:05.280 align:middle line:84% with the pigeonhole principle, you generally 00:31:05.280 --> 00:31:07.500 align:middle line:84% need to be very careful about what your pigeons are, 00:31:07.500 --> 00:31:10.040 align:middle line:84% what your pigeonholes are, and what the map between them is. 00:31:10.040 --> 00:31:14.791 align:middle line:90% 00:31:14.791 --> 00:31:17.125 align:middle line:84% So what are those three elements in this case? 00:31:17.125 --> 00:31:29.840 align:middle line:90% 00:31:29.840 --> 00:31:34.970 align:middle line:84% Maybe we could phrase it instead as, if I have n plus 1, 00:31:34.970 --> 00:31:36.310 align:middle line:90% then it guarantees a pair. 00:31:36.310 --> 00:31:43.790 align:middle line:90% 00:31:43.790 --> 00:31:45.470 align:middle line:90% Yeah. 00:31:45.470 --> 00:31:47.720 align:middle line:90% AUDIENCE: And the [INAUDIBLE] 00:31:47.720 --> 00:31:53.231 align:middle line:90% 00:31:53.231 --> 00:31:59.030 align:middle line:84% BRYNMOR CHAPMAN: So A is the set of socks, 00:31:59.030 --> 00:32:01.403 align:middle line:90% and B is a set of colors. 00:32:01.403 --> 00:32:05.090 align:middle line:90% 00:32:05.090 --> 00:32:08.725 align:middle line:84% And the map is going to map the sock to its color. 00:32:08.725 --> 00:32:17.610 align:middle line:90% 00:32:17.610 --> 00:32:19.430 align:middle line:84% Does that make sense to everybody? 00:32:19.430 --> 00:32:27.957 align:middle line:90% 00:32:27.957 --> 00:32:30.555 align:middle line:84% Does anybody know what the population of Boston is? 00:32:30.555 --> 00:32:34.570 align:middle line:90% 00:32:34.570 --> 00:32:35.820 align:middle line:90% AUDIENCE: [INAUDIBLE] 00:32:35.820 --> 00:32:38.070 align:middle line:84% BRYNMOR CHAPMAN: Are you looking at the lecture notes? 00:32:38.070 --> 00:32:39.070 align:middle line:90% AUDIENCE: Yeah. 00:32:39.070 --> 00:32:40.720 align:middle line:90% BRYNMOR CHAPMAN: Oh, OK. 00:32:40.720 --> 00:32:44.680 align:middle line:84% Yeah, about 650,000 as of however many decades ago these 00:32:44.680 --> 00:32:46.810 align:middle line:90% lecture notes were written. 00:32:46.810 --> 00:32:48.800 align:middle line:84% I don't actually know that was very long ago. 00:32:48.800 --> 00:32:50.055 align:middle line:90% It could be current. 00:32:50.055 --> 00:33:04.060 align:middle line:90% 00:33:04.060 --> 00:33:07.640 align:middle line:90% Is anybody a human biology nerd? 00:33:07.640 --> 00:33:09.575 align:middle line:84% Does anybody know how many hairs humans have? 00:33:09.575 --> 00:33:18.360 align:middle line:90% 00:33:18.360 --> 00:33:30.120 align:middle line:84% So it turns out humans have less than about 200,000 hairs. 00:33:30.120 --> 00:33:32.525 align:middle line:84% So what does the pigeonhole principle tell us here? 00:33:32.525 --> 00:33:38.670 align:middle line:90% 00:33:38.670 --> 00:33:39.485 align:middle line:90% Any ideas? 00:33:39.485 --> 00:33:48.270 align:middle line:90% 00:33:48.270 --> 00:33:49.420 align:middle line:90% Getting some facepalms. 00:33:49.420 --> 00:33:50.624 align:middle line:90% Yeah. 00:33:50.624 --> 00:33:52.980 align:middle line:84% AUDIENCE: That at least three humans 00:33:52.980 --> 00:33:54.480 align:middle line:90% have the same number of hairs. 00:33:54.480 --> 00:33:55.840 align:middle line:90% BRYNMOR CHAPMAN: Oh, good point. 00:33:55.840 --> 00:33:59.070 align:middle line:84% So the answer was at least three humans in Boston 00:33:59.070 --> 00:34:00.320 align:middle line:90% have the same number of hairs. 00:34:00.320 --> 00:34:15.489 align:middle line:90% 00:34:15.489 --> 00:34:18.810 align:middle line:84% So as stated, all we know is that at least two Bostonians 00:34:18.810 --> 00:34:21.199 align:middle line:90% have the same number of hairs. 00:34:21.199 --> 00:34:23.699 align:middle line:84% So does that mean that there's somebody with the same number 00:34:23.699 --> 00:34:24.515 align:middle line:90% of hairs as me? 00:34:24.515 --> 00:34:28.110 align:middle line:90% 00:34:28.110 --> 00:34:30.556 align:middle line:90% Not necessarily. 00:34:30.556 --> 00:34:32.980 align:middle line:84% So we know that there exists two Bostonians. 00:34:32.980 --> 00:34:35.266 align:middle line:90% We don't know who they are. 00:34:35.266 --> 00:34:37.540 align:middle line:90% This is non-constructive. 00:34:37.540 --> 00:34:40.260 align:middle line:90% It only tells us of existence. 00:34:40.260 --> 00:34:44.639 align:middle line:84% But your colleague also brings up another good point. 00:34:44.639 --> 00:34:48.880 align:middle line:84% Because this 650,000 is so much larger than the 200,000, 00:34:48.880 --> 00:34:51.670 align:middle line:84% we can actually say something stronger. 00:34:51.670 --> 00:34:55.510 align:middle line:90% All we need is 200,000 plus 1. 00:34:55.510 --> 00:35:00.600 align:middle line:84% That'll give us a collision of two Bostonians. 00:35:00.600 --> 00:35:04.620 align:middle line:84% We've actually got more than three times that number. 00:35:04.620 --> 00:35:07.600 align:middle line:84% So there are more than three Bostonians who 00:35:07.600 --> 00:35:08.850 align:middle line:90% have the same number of hairs. 00:35:08.850 --> 00:35:10.433 align:middle line:84% So in fact, we can conclude that there 00:35:10.433 --> 00:35:13.790 align:middle line:84% are at least four Bostonians who all 00:35:13.790 --> 00:35:15.040 align:middle line:90% have the same number of hairs. 00:35:15.040 --> 00:35:20.372 align:middle line:90% 00:35:20.372 --> 00:35:22.140 align:middle line:84% Does that make sense to everybody? 00:35:22.140 --> 00:35:23.000 align:middle line:90% Yeah. 00:35:23.000 --> 00:35:25.082 align:middle line:84% AUDIENCE: How did you get the four [INAUDIBLE]? 00:35:25.082 --> 00:35:26.790 align:middle line:84% BRYNMOR CHAPMAN: Ah, so the question was, 00:35:26.790 --> 00:35:28.080 align:middle line:90% how did we get the four? 00:35:28.080 --> 00:35:30.590 align:middle line:84% So this is what we call the generalized pigeonhole 00:35:30.590 --> 00:35:31.560 align:middle line:90% principle. 00:35:31.560 --> 00:35:40.682 align:middle line:90% 00:35:40.682 --> 00:35:47.510 align:middle line:84% So up here, we just said, if A is larger than B, 00:35:47.510 --> 00:35:53.000 align:middle line:84% but if we have the stronger condition that if A is larger 00:35:53.000 --> 00:36:00.710 align:middle line:84% than k times B, then we've got a k-plus-1-way collision, 00:36:00.710 --> 00:36:02.340 align:middle line:90% not just a two-way collision. 00:36:02.340 --> 00:36:25.605 align:middle line:90% 00:36:25.605 --> 00:36:31.660 align:middle line:84% So because we have 650,000, which is more than 3 times 00:36:31.660 --> 00:36:34.870 align:middle line:84% 200,000, we've got a 3-plus-1 way collision, 00:36:34.870 --> 00:36:36.350 align:middle line:90% not just a three-way collision. 00:36:36.350 --> 00:36:37.930 align:middle line:90% Does that make sense? 00:36:37.930 --> 00:36:38.890 align:middle line:90% Yeah. 00:36:38.890 --> 00:36:41.630 align:middle line:84% AUDIENCE: So [INAUDIBLE] in this example, 00:36:41.630 --> 00:36:44.440 align:middle line:84% the number of possible hairs you might have [INAUDIBLE] 00:36:44.440 --> 00:36:47.930 align:middle line:84% every integer or natural number from 1 to [INAUDIBLE]? 00:36:47.930 --> 00:36:49.090 align:middle line:90% BRYNMOR CHAPMAN: Yeah. 00:36:49.090 --> 00:36:50.650 align:middle line:84% I suppose we could include 0, but I 00:36:50.650 --> 00:36:53.450 align:middle line:84% guess everybody knows that there are bald Bostonians. 00:36:53.450 --> 00:36:55.930 align:middle line:90% So maybe we exclude that. 00:36:55.930 --> 00:37:00.160 align:middle line:84% But yeah, every natural number from 1 up to 200,000 is going 00:37:00.160 --> 00:37:04.600 align:middle line:84% to be our B, and our A is going to be the set of all Bostonians. 00:37:04.600 --> 00:37:06.865 align:middle line:84% And we are mapping a person to the number of hairs 00:37:06.865 --> 00:37:07.490 align:middle line:90% that they have. 00:37:07.490 --> 00:37:15.280 align:middle line:90% 00:37:15.280 --> 00:37:20.640 align:middle line:84% So you can basically always use the generalized pigeonhole 00:37:20.640 --> 00:37:22.177 align:middle line:90% principle if you would-- 00:37:22.177 --> 00:37:23.760 align:middle line:84% just like strong induction will always 00:37:23.760 --> 00:37:26.880 align:middle line:84% work whenever weak induction does. 00:37:26.880 --> 00:37:28.545 align:middle line:90% You could always set k equal 1. 00:37:28.545 --> 00:37:33.550 align:middle line:90% 00:37:33.550 --> 00:37:36.480 align:middle line:84% But the proof is basically the same, 00:37:36.480 --> 00:37:41.160 align:middle line:84% and if you use these to prove things about functions, 00:37:41.160 --> 00:37:45.900 align:middle line:84% the proofs can often be very short, very simple. 00:37:45.900 --> 00:37:50.230 align:middle line:84% But they can also sometimes require some cleverness. 00:37:50.230 --> 00:37:57.050 align:middle line:84% So it may not always be obvious what A, B, and f should be. 00:37:57.050 --> 00:37:59.033 align:middle line:84% So here's a slightly more involved example. 00:37:59.033 --> 00:38:11.430 align:middle line:90% 00:38:11.430 --> 00:38:22.525 align:middle line:84% So we have an 8 by 8 chessboard and 33 rooks. 00:38:22.525 --> 00:38:25.670 align:middle line:90% 00:38:25.670 --> 00:38:31.080 align:middle line:84% Does everybody know how chess works, vaguely? 00:38:31.080 --> 00:38:36.330 align:middle line:84% So rooks can move along ranks or files, ranks or files. 00:38:36.330 --> 00:38:38.810 align:middle line:90% Sorry. 00:38:38.810 --> 00:38:48.868 align:middle line:84% So we would like to place these rooks so that any set of five-- 00:38:48.868 --> 00:38:52.610 align:middle line:84% they don't all attack each other. 00:38:52.610 --> 00:38:56.540 align:middle line:90% Or no, I phrased that wrong. 00:38:56.540 --> 00:39:00.140 align:middle line:84% So we want to find five of them where none of them 00:39:00.140 --> 00:39:01.700 align:middle line:90% attack each other. 00:39:01.700 --> 00:39:03.370 align:middle line:84% So basically, we want them all to be 00:39:03.370 --> 00:39:05.780 align:middle line:90% in distinct rows and files. 00:39:05.780 --> 00:39:13.460 align:middle line:84% So if we've got something like this, if we have a rook here, 00:39:13.460 --> 00:39:16.060 align:middle line:84% a rook here, and a rook here, those three 00:39:16.060 --> 00:39:18.940 align:middle line:90% do not attack each other. 00:39:18.940 --> 00:39:23.020 align:middle line:84% So we're trying to find a set of five like that. 00:39:23.020 --> 00:39:26.547 align:middle line:84% No two of them share a column or a row. 00:39:26.547 --> 00:39:27.630 align:middle line:90% Does the setup make sense? 00:39:27.630 --> 00:39:31.370 align:middle line:90% 00:39:31.370 --> 00:39:32.295 align:middle line:90% How could we do that? 00:39:32.295 --> 00:39:36.160 align:middle line:90% 00:39:36.160 --> 00:39:38.580 align:middle line:84% Rather, how could we prove the existence of such a set? 00:39:38.580 --> 00:39:48.010 align:middle line:90% 00:39:48.010 --> 00:39:51.520 align:middle line:84% Maybe it would help to draw out the chessboard. 00:39:51.520 --> 00:39:52.350 align:middle line:90% Let's see. 00:39:52.350 --> 00:39:55.690 align:middle line:90% 00:39:55.690 --> 00:40:14.460 align:middle line:84% 1, 2, 3, 4, 5, 6, 7, 1, 2, 3, 4, 5, 6, 7. 00:40:14.460 --> 00:40:16.210 align:middle line:90% There's our wonky chessboard. 00:40:16.210 --> 00:40:40.210 align:middle line:90% 00:40:40.210 --> 00:40:41.820 align:middle line:90% So how could we do this? 00:40:41.820 --> 00:40:43.650 align:middle line:90% Any ideas? 00:40:43.650 --> 00:40:44.580 align:middle line:90% Yeah. 00:40:44.580 --> 00:40:51.670 align:middle line:84% AUDIENCE: Maybe [INAUDIBLE] the five pieces. 00:40:51.670 --> 00:40:56.610 align:middle line:84% So that's 32 [INAUDIBLE] we have 33 [INAUDIBLE]. 00:40:56.610 --> 00:41:00.390 align:middle line:84% BRYNMOR CHAPMAN: So the answer is, if we place four rooks, 00:41:00.390 --> 00:41:03.053 align:middle line:84% they cover at most 2 to the 5 squares? 00:41:03.053 --> 00:41:08.000 align:middle line:90% 00:41:08.000 --> 00:41:08.733 align:middle line:90% Is that true? 00:41:08.733 --> 00:41:11.570 align:middle line:90% 00:41:11.570 --> 00:41:13.110 align:middle line:90% I don't think that's true. 00:41:13.110 --> 00:41:17.160 align:middle line:84% If we place rooks in these four squares, 00:41:17.160 --> 00:41:21.680 align:middle line:84% they're going to cover 3/4 of the chessboard. 00:41:21.680 --> 00:41:22.730 align:middle line:90% Yes. 00:41:22.730 --> 00:41:24.170 align:middle line:90% AUDIENCE: I'm just curious. 00:41:24.170 --> 00:41:27.060 align:middle line:84% Do the [INAUDIBLE] have to not attack each other, 00:41:27.060 --> 00:41:30.430 align:middle line:84% but their position relevant to the other 28 doesn't matter? 00:41:30.430 --> 00:41:31.680 align:middle line:90% BRYNMOR CHAPMAN: That's right. 00:41:31.680 --> 00:41:32.510 align:middle line:90% AUDIENCE: OK. 00:41:32.510 --> 00:41:34.220 align:middle line:90% This is a simple problem. 00:41:34.220 --> 00:41:36.090 align:middle line:90% BRYNMOR CHAPMAN: Pardon? 00:41:36.090 --> 00:41:39.140 align:middle line:84% AUDIENCE: If there's only five that need to be distinct 00:41:39.140 --> 00:41:43.130 align:middle line:84% and we have eight rows and eight columns, 00:41:43.130 --> 00:41:48.410 align:middle line:84% then it's pretty easy to determine. 00:41:48.410 --> 00:41:50.450 align:middle line:84% BRYNMOR CHAPMAN: So what is your solution then? 00:41:50.450 --> 00:41:54.600 align:middle line:84% AUDIENCE: You can literally just place five of them diagonally, 00:41:54.600 --> 00:41:55.220 align:middle line:90% couldn't you? 00:41:55.220 --> 00:41:56.595 align:middle line:84% BRYNMOR CHAPMAN: Oh, oh, sorry, I 00:41:56.595 --> 00:41:58.750 align:middle line:84% think you may have misunderstood the problem. 00:41:58.750 --> 00:42:01.490 align:middle line:90% We have 33 rocks. 00:42:01.490 --> 00:42:03.250 align:middle line:84% We're trying to find a set of five that 00:42:03.250 --> 00:42:04.250 align:middle line:90% don't attack each other. 00:42:04.250 --> 00:42:05.968 align:middle line:90% We're given the placement. 00:42:05.968 --> 00:42:07.760 align:middle line:84% We're not allowed to just place five rocks. 00:42:07.760 --> 00:42:08.870 align:middle line:84% If we could place them, then yeah, 00:42:08.870 --> 00:42:10.630 align:middle line:84% we could place them along a diagonal, 00:42:10.630 --> 00:42:13.020 align:middle line:90% and then we've got eight. 00:42:13.020 --> 00:42:15.430 align:middle line:90% So some of these have rooks. 00:42:15.430 --> 00:42:16.690 align:middle line:90% We don't get to control where. 00:42:16.690 --> 00:42:19.750 align:middle line:90% 00:42:19.750 --> 00:42:24.250 align:middle line:84% And we're trying to find some subset of these, 00:42:24.250 --> 00:42:28.450 align:middle line:84% some subset of five of these that don't attack each other. 00:42:28.450 --> 00:42:31.360 align:middle line:84% Does the problem setup make sense? 00:42:31.360 --> 00:42:32.200 align:middle line:90% Maybe? 00:42:32.200 --> 00:42:32.780 align:middle line:90% Yes? 00:42:32.780 --> 00:42:33.625 align:middle line:90% No? 00:42:33.625 --> 00:42:34.125 align:middle line:90% Yeah? 00:42:34.125 --> 00:42:36.647 align:middle line:90% 00:42:36.647 --> 00:42:38.730 align:middle line:84% So how could we use the pigeonhole principle here? 00:42:38.730 --> 00:42:46.120 align:middle line:90% 00:42:46.120 --> 00:42:48.350 align:middle line:84% Well, perhaps let's start with the easiest element. 00:42:48.350 --> 00:42:49.575 align:middle line:90% What should our pigeons be? 00:42:49.575 --> 00:42:54.100 align:middle line:90% 00:42:54.100 --> 00:42:56.070 align:middle line:84% What are we trying to make collide? 00:42:56.070 --> 00:42:56.740 align:middle line:90% Yeah. 00:42:56.740 --> 00:42:57.525 align:middle line:90% AUDIENCE: Rooks. 00:42:57.525 --> 00:42:57.900 align:middle line:90% BRYNMOR CHAPMAN: Rooks. 00:42:57.900 --> 00:42:58.420 align:middle line:90% Yeah. 00:42:58.420 --> 00:43:00.610 align:middle line:84% So our pigeons are going to be the rooks. 00:43:00.610 --> 00:43:01.610 align:middle line:90% And I've lost my eraser. 00:43:01.610 --> 00:43:20.370 align:middle line:90% 00:43:20.370 --> 00:43:24.065 align:middle line:84% Now, what should our pigeonholes be? 00:43:24.065 --> 00:43:27.990 align:middle line:90% 00:43:27.990 --> 00:43:33.720 align:middle line:84% We want to set it up so that if we get a five-way collision, 00:43:33.720 --> 00:43:36.065 align:middle line:84% those five rooks do not attack each other. 00:43:36.065 --> 00:43:41.550 align:middle line:90% 00:43:41.550 --> 00:43:42.547 align:middle line:90% Yeah. 00:43:42.547 --> 00:43:45.130 align:middle line:84% AUDIENCE: [INAUDIBLE] the rooks are being placed on the board, 00:43:45.130 --> 00:43:47.880 align:middle line:84% and the pigeonholes should be the squares that 00:43:47.880 --> 00:43:48.883 align:middle line:90% are on the chessboard. 00:43:48.883 --> 00:43:49.800 align:middle line:90% BRYNMOR CHAPMAN: Yeah. 00:43:49.800 --> 00:43:52.720 align:middle line:84% So the answer was, we're placing these rooks on the chessboard, 00:43:52.720 --> 00:43:55.780 align:middle line:84% so let's make the pigeonholes the squares of the chessboard. 00:43:55.780 --> 00:44:06.930 align:middle line:90% 00:44:06.930 --> 00:44:09.190 align:middle line:90% So you are off to a good start. 00:44:09.190 --> 00:44:10.680 align:middle line:90% That's a good idea. 00:44:10.680 --> 00:44:13.800 align:middle line:84% The problem right now is that we've got fewer rooks 00:44:13.800 --> 00:44:16.750 align:middle line:90% than squares on the chessboard. 00:44:16.750 --> 00:44:21.220 align:middle line:84% So what we really want is a five-way collision, 00:44:21.220 --> 00:44:26.190 align:middle line:84% so we want to ensure that we have more than four times 00:44:26.190 --> 00:44:30.840 align:middle line:90% as many rooks as pigeonholes. 00:44:30.840 --> 00:44:36.030 align:middle line:84% So is there a way that we can modify your solution 00:44:36.030 --> 00:44:39.695 align:middle line:84% to make fewer things where collision is still good for us? 00:44:39.695 --> 00:44:45.390 align:middle line:90% 00:44:45.390 --> 00:44:50.102 align:middle line:84% So can we subdivide our squares so that we have-- yeah. 00:44:50.102 --> 00:44:51.790 align:middle line:90% AUDIENCE: [INAUDIBLE] 00:44:51.790 --> 00:44:53.116 align:middle line:90% BRYNMOR CHAPMAN: Sorry? 00:44:53.116 --> 00:44:54.980 align:middle line:90% AUDIENCE: [INAUDIBLE] 00:44:54.980 --> 00:44:58.570 align:middle line:84% BRYNMOR CHAPMAN: The answer was, squares that don't have rooks. 00:44:58.570 --> 00:45:00.640 align:middle line:84% That's not going to be very helpful. 00:45:00.640 --> 00:45:02.210 align:middle line:84% We're assigning rooks to pigeonholes. 00:45:02.210 --> 00:45:10.900 align:middle line:84% So the map should map rooks to occupied squares of some sort. 00:45:10.900 --> 00:45:13.270 align:middle line:84% But how could we maybe split up the chessboard 00:45:13.270 --> 00:45:15.200 align:middle line:90% into eight pigeonholes? 00:45:15.200 --> 00:45:17.840 align:middle line:84% If we have eight pigeonholes, 33 rooks, 00:45:17.840 --> 00:45:19.640 align:middle line:84% that's more than four times as many. 00:45:19.640 --> 00:45:22.600 align:middle line:90% So we get a five-way collision. 00:45:22.600 --> 00:45:25.120 align:middle line:84% What could our eight sets of squares be? 00:45:25.120 --> 00:45:25.900 align:middle line:90% Yeah. 00:45:25.900 --> 00:45:26.650 align:middle line:90% AUDIENCE: Columns. 00:45:26.650 --> 00:45:28.760 align:middle line:90% BRYNMOR CHAPMAN: Columns. 00:45:28.760 --> 00:45:32.025 align:middle line:84% So if we have five rooks in a column, is that helpful? 00:45:32.025 --> 00:45:37.600 align:middle line:90% 00:45:37.600 --> 00:45:41.170 align:middle line:84% All of those rooks attack each other then. 00:45:41.170 --> 00:45:42.520 align:middle line:90% You're getting close, though. 00:45:42.520 --> 00:45:43.780 align:middle line:90% Yeah. 00:45:43.780 --> 00:45:47.080 align:middle line:84% AUDIENCE: It'd be like diagonals [INAUDIBLE] offset by one. 00:45:47.080 --> 00:45:48.190 align:middle line:90% BRYNMOR CHAPMAN: Yeah. 00:45:48.190 --> 00:45:53.890 align:middle line:84% So the answer was diagonals but then offset them. 00:45:53.890 --> 00:45:56.610 align:middle line:84% So I'm going to number of these squares. 00:45:56.610 --> 00:45:59.160 align:middle line:84% So everything numbered 1 is going 00:45:59.160 --> 00:46:00.990 align:middle line:90% to be our first pigeonhole. 00:46:00.990 --> 00:46:05.537 align:middle line:84% So if we get five rooks along the main diagonal, all of them 00:46:05.537 --> 00:46:07.870 align:middle line:84% are going to be in different rows and different columns, 00:46:07.870 --> 00:46:09.220 align:middle line:90% so they won't attack each other. 00:46:09.220 --> 00:46:11.790 align:middle line:90% So that's good. 00:46:11.790 --> 00:46:17.190 align:middle line:84% Number 2 is going to be offset by one, 00:46:17.190 --> 00:46:21.150 align:middle line:84% but then we're also going to wrap around down here. 00:46:21.150 --> 00:46:24.030 align:middle line:84% So once again, we've got eight squares 00:46:24.030 --> 00:46:26.970 align:middle line:84% that are all in different rows and different columns. 00:46:26.970 --> 00:46:30.180 align:middle line:84% So if we get five on two squares, 00:46:30.180 --> 00:46:32.220 align:middle line:84% they also don't attack each other. 00:46:32.220 --> 00:46:34.440 align:middle line:90% And we can continue in this way. 00:46:34.440 --> 00:46:40.835 align:middle line:84% Three goes along here and then these two squares, et cetera. 00:46:40.835 --> 00:46:44.388 align:middle line:90% 00:46:44.388 --> 00:46:50.180 align:middle line:84% So now we have eight pigeonholes, 33 rooks, 00:46:50.180 --> 00:46:52.050 align:middle line:84% so that's more than four times as many. 00:46:52.050 --> 00:46:55.100 align:middle line:90% We get a five-way collision. 00:46:55.100 --> 00:46:59.082 align:middle line:84% Whichever number of square has that five-way collision, 00:46:59.082 --> 00:47:01.165 align:middle line:84% those are five rooks that don't attack each other. 00:47:01.165 --> 00:47:11.191 align:middle line:90% 00:47:11.191 --> 00:47:14.450 align:middle line:84% Does that make sense to everybody? 00:47:14.450 --> 00:47:16.130 align:middle line:84% So it took a little bit of creativity 00:47:16.130 --> 00:47:22.520 align:middle line:84% to come up with the pigeonholes here, but once we have that, 00:47:22.520 --> 00:47:24.390 align:middle line:84% the proof is fairly straightforward. 00:47:24.390 --> 00:47:27.020 align:middle line:90% 00:47:27.020 --> 00:47:29.870 align:middle line:84% So let's change gears a little bit and move 00:47:29.870 --> 00:47:34.550 align:middle line:84% on to the last counting technique 00:47:34.550 --> 00:47:46.340 align:middle line:84% for today, which we call combinatorial proofs or double 00:47:46.340 --> 00:47:47.625 align:middle line:90% counting. 00:47:47.625 --> 00:47:48.125 align:middle line:90% Oops. 00:47:48.125 --> 00:47:57.710 align:middle line:90% 00:47:57.710 --> 00:48:01.520 align:middle line:84% So I'd like to think of this as somewhat similar to-- 00:48:01.520 --> 00:48:04.890 align:middle line:84% I believe you've seen the bijection rule earlier. 00:48:04.890 --> 00:48:07.370 align:middle line:84% So if you have two sets and you can come up 00:48:07.370 --> 00:48:09.680 align:middle line:84% with a bijection between them, then those two sets 00:48:09.680 --> 00:48:10.465 align:middle line:90% are the same size. 00:48:10.465 --> 00:48:13.130 align:middle line:90% 00:48:13.130 --> 00:48:15.530 align:middle line:84% Well, similarly, if you have two sizes 00:48:15.530 --> 00:48:18.750 align:middle line:84% in a single set that shares those sizes, 00:48:18.750 --> 00:48:20.900 align:middle line:90% those sizes must be equal. 00:48:20.900 --> 00:48:26.690 align:middle line:84% So basically, this is a proof technique where we take a set 00:48:26.690 --> 00:48:29.330 align:middle line:84% and count it in two different ways, 00:48:29.330 --> 00:48:33.980 align:middle line:84% and that can give us some nice combinatorial identities. 00:48:33.980 --> 00:48:49.090 align:middle line:84% So for instance, we have this identity. 00:48:49.090 --> 00:48:52.030 align:middle line:84% Have people seen this one before? 00:48:52.030 --> 00:48:54.230 align:middle line:90% No? 00:48:54.230 --> 00:48:59.030 align:middle line:84% So if you sum up n choose k over all k from 0 to n, 00:48:59.030 --> 00:49:01.555 align:middle line:90% you just get 2 to the n. 00:49:01.555 --> 00:49:02.930 align:middle line:84% So if you haven't seen it before, 00:49:02.930 --> 00:49:05.830 align:middle line:84% it's perhaps a little bit counterintuitive. 00:49:05.830 --> 00:49:10.670 align:middle line:84% We've got a summation of a bunch of things involving factorials, 00:49:10.670 --> 00:49:15.472 align:middle line:84% but it turns out to be some nice exponential. 00:49:15.472 --> 00:49:19.360 align:middle line:90% So how could we prove this? 00:49:19.360 --> 00:49:23.500 align:middle line:84% How could we come up with a set that we 00:49:23.500 --> 00:49:27.490 align:middle line:84% can count both as this expression 00:49:27.490 --> 00:49:31.240 align:middle line:90% and as this expression? 00:49:31.240 --> 00:49:33.100 align:middle line:90% Any ideas? 00:49:33.100 --> 00:49:34.570 align:middle line:90% Yeah. 00:49:34.570 --> 00:49:36.900 align:middle line:84% AUDIENCE: [INAUDIBLE] any length n binary sequence. 00:49:36.900 --> 00:49:40.930 align:middle line:90% 00:49:40.930 --> 00:49:43.840 align:middle line:84% BRYNMOR CHAPMAN: So the solution was count length n binary 00:49:43.840 --> 00:49:44.560 align:middle line:90% sequences. 00:49:44.560 --> 00:49:59.040 align:middle line:90% 00:49:59.040 --> 00:50:03.390 align:middle line:84% So how does that give you each side of the equality? 00:50:03.390 --> 00:50:06.300 align:middle line:84% AUDIENCE: On the right side, [INAUDIBLE] 00:50:06.300 --> 00:50:08.250 align:middle line:90% a 1 or a 0 in each position. 00:50:08.250 --> 00:50:11.290 align:middle line:84% [INAUDIBLE] can imagine, well, first of all, 00:50:11.290 --> 00:50:14.040 align:middle line:84% how many are there that have 1 and 1 [INAUDIBLE] 00:50:14.040 --> 00:50:16.050 align:middle line:90% be like, choose one. 00:50:16.050 --> 00:50:18.450 align:middle line:84% How many others have [INAUDIBLE] choose two. 00:50:18.450 --> 00:50:20.640 align:middle line:84% And you have to consider all sequences that 00:50:20.640 --> 00:50:23.130 align:middle line:84% have any number of 1's in there [INAUDIBLE] 00:50:23.130 --> 00:50:26.130 align:middle line:90% 00:50:26.130 --> 00:50:27.690 align:middle line:90% BRYNMOR CHAPMAN: Exactly. 00:50:27.690 --> 00:50:28.300 align:middle line:90% Yeah. 00:50:28.300 --> 00:50:33.630 align:middle line:84% So the right-hand side is fairly straightforward. 00:50:33.630 --> 00:50:35.040 align:middle line:84% If we're counting length n binary 00:50:35.040 --> 00:50:39.460 align:middle line:84% sequences, every position, you can either have 0 or 1. 00:50:39.460 --> 00:50:42.020 align:middle line:84% So product rule gives us 2 to the n. 00:50:42.020 --> 00:50:48.960 align:middle line:90% 00:50:48.960 --> 00:50:52.950 align:middle line:84% And for the left-hand side, we're 00:50:52.950 --> 00:50:56.070 align:middle line:84% going to partition based on how many 1's 00:50:56.070 --> 00:50:58.950 align:middle line:90% there are in our sequence. 00:50:58.950 --> 00:51:02.280 align:middle line:84% So our sequence could have any number 00:51:02.280 --> 00:51:05.410 align:middle line:90% of 1's from up to n inclusive. 00:51:05.410 --> 00:51:07.920 align:middle line:90% 00:51:07.920 --> 00:51:09.670 align:middle line:90% And all of these are disjoint. 00:51:09.670 --> 00:51:13.230 align:middle line:84% If it has three 1's, it can't also have eight 1's. 00:51:13.230 --> 00:51:16.680 align:middle line:84% So by our sum rule, we can just sum up 00:51:16.680 --> 00:51:20.190 align:middle line:90% all of the sizes of those sets. 00:51:20.190 --> 00:51:24.450 align:middle line:84% What is the size of the set of sequences that has, say, 00:51:24.450 --> 00:51:25.355 align:middle line:90% three 1's? 00:51:25.355 --> 00:51:34.740 align:middle line:90% 00:51:34.740 --> 00:51:37.500 align:middle line:90% Anybody? 00:51:37.500 --> 00:51:40.295 align:middle line:84% How many binary sequences of length n have three 1's? 00:51:40.295 --> 00:51:50.650 align:middle line:90% 00:51:50.650 --> 00:51:51.607 align:middle line:90% AUDIENCE: n choose 3. 00:51:51.607 --> 00:51:52.940 align:middle line:90% BRYNMOR CHAPMAN: n choose three. 00:51:52.940 --> 00:51:53.800 align:middle line:90% Yeah. 00:51:53.800 --> 00:52:07.420 align:middle line:84% Or more generally, there are n choose k with k 1's, and k 00:52:07.420 --> 00:52:15.600 align:middle line:84% from 0 to n form a partition of our set. 00:52:15.600 --> 00:52:19.960 align:middle line:90% 00:52:19.960 --> 00:52:24.520 align:middle line:84% So if we sum up all of these binomial coefficients from 0 00:52:24.520 --> 00:52:30.130 align:middle line:84% to n, we're also going to get the number of length n binary 00:52:30.130 --> 00:52:32.920 align:middle line:90% sequences. 00:52:32.920 --> 00:52:34.700 align:middle line:84% Does that make sense to everybody? 00:52:34.700 --> 00:52:38.390 align:middle line:90% 00:52:38.390 --> 00:52:40.600 align:middle line:84% Does the proof technique make sense to everybody? 00:52:40.600 --> 00:52:43.710 align:middle line:84% We're taking a single set, and we're counting it 00:52:43.710 --> 00:52:46.470 align:middle line:90% in two different ways. 00:52:46.470 --> 00:52:50.410 align:middle line:84% And each way of counting gives us a different expression. 00:52:50.410 --> 00:52:52.065 align:middle line:84% So those two expressions must be equal. 00:52:52.065 --> 00:52:56.250 align:middle line:90% 00:52:56.250 --> 00:52:59.795 align:middle line:84% So the key here is choosing the right set to count. 00:52:59.795 --> 00:53:18.000 align:middle line:90% 00:53:18.000 --> 00:53:21.150 align:middle line:84% So as it turns out, that's actually a special case 00:53:21.150 --> 00:53:25.710 align:middle line:90% of a nice theorem. 00:53:25.710 --> 00:53:28.660 align:middle line:84% So we've been using binomial coefficients quite a lot, 00:53:28.660 --> 00:53:31.990 align:middle line:90% these n choose k terms. 00:53:31.990 --> 00:53:33.395 align:middle line:90% Does anybody know their eponym? 00:53:33.395 --> 00:53:37.530 align:middle line:90% 00:53:37.530 --> 00:53:40.128 align:middle line:90% Any ideas? 00:53:40.128 --> 00:53:42.295 align:middle line:84% Has anybody encountered the binomial theorem before? 00:53:42.295 --> 00:53:45.170 align:middle line:90% 00:53:45.170 --> 00:53:46.005 align:middle line:90% No? 00:53:46.005 --> 00:53:47.005 align:middle line:90% Well, let's take a look. 00:53:47.005 --> 00:54:10.050 align:middle line:90% 00:54:10.050 --> 00:54:14.675 align:middle line:84% So this here is what we call a binomial, 00:54:14.675 --> 00:54:19.190 align:middle line:90% a sum of two monomials. 00:54:19.190 --> 00:54:21.950 align:middle line:84% And the binomial theorem says that if we take a binomial, 00:54:21.950 --> 00:54:27.230 align:middle line:84% raise it to the n-th power, then we get basically the same thing 00:54:27.230 --> 00:54:29.070 align:middle line:84% as we had on the left-hand side over there, 00:54:29.070 --> 00:54:33.055 align:middle line:84% but we've got these powers of x and y. 00:54:33.055 --> 00:54:37.304 align:middle line:90% 00:54:37.304 --> 00:54:41.080 align:middle line:90% And by convention, 0 to the 0-- 00:54:41.080 --> 00:54:42.630 align:middle line:84% we just say that that's equal to 1. 00:54:42.630 --> 00:54:48.130 align:middle line:90% 00:54:48.130 --> 00:55:01.720 align:middle line:84% So the proof-- does anybody know how we might prove this? 00:55:01.720 --> 00:55:03.465 align:middle line:84% It's fairly similar to what we just did. 00:55:03.465 --> 00:55:14.530 align:middle line:90% 00:55:14.530 --> 00:55:17.535 align:middle line:84% Well, what happens if we expand out the left-hand side? 00:55:17.535 --> 00:55:25.090 align:middle line:90% 00:55:25.090 --> 00:55:25.965 align:middle line:90% What do we get? 00:55:25.965 --> 00:55:31.510 align:middle line:90% 00:55:31.510 --> 00:55:32.740 align:middle line:90% Yeah. 00:55:32.740 --> 00:55:36.413 align:middle line:84% AUDIENCE: You get coefficients of Pascal's triangle 00:55:36.413 --> 00:55:37.315 align:middle line:90% [INAUDIBLE]. 00:55:37.315 --> 00:55:39.462 align:middle line:90% 00:55:39.462 --> 00:55:40.920 align:middle line:84% BRYNMOR CHAPMAN: So the answer was, 00:55:40.920 --> 00:55:43.830 align:middle line:84% we got the coefficients of Pascal's triangle. 00:55:43.830 --> 00:55:46.290 align:middle line:84% You've skipped a couple of steps. 00:55:46.290 --> 00:55:47.870 align:middle line:84% More immediately, what do we get? 00:55:47.870 --> 00:55:53.550 align:middle line:90% 00:55:53.550 --> 00:55:57.570 align:middle line:84% Say, if n equals 2, what do we get? 00:55:57.570 --> 00:56:01.540 align:middle line:84% We get xx plus xy plus yx plus yy. 00:56:01.540 --> 00:56:05.300 align:middle line:84% We've basically got a bunch of binary sequences of x's and y's. 00:56:05.300 --> 00:56:08.935 align:middle line:90% 00:56:08.935 --> 00:56:18.663 align:middle line:84% So this gives us a sum of 2 to the n terms of the form. 00:56:18.663 --> 00:56:22.530 align:middle line:90% 00:56:22.530 --> 00:56:29.160 align:middle line:84% And I'll call it b1, b2, up to bn 00:56:29.160 --> 00:56:35.980 align:middle line:90% where each b is either x or y. 00:56:35.980 --> 00:56:48.360 align:middle line:90% 00:56:48.360 --> 00:56:52.250 align:middle line:84% So now how do we get Pascal out of that? 00:56:52.250 --> 00:56:59.840 align:middle line:84% AUDIENCE: You write [INAUDIBLE] the triangle 00:56:59.840 --> 00:57:03.020 align:middle line:84% as a different value [INAUDIBLE] start with the coefficient 00:57:03.020 --> 00:57:07.110 align:middle line:90% of the first [INAUDIBLE]. 00:57:07.110 --> 00:57:09.360 align:middle line:84% BRYNMOR CHAPMAN: So you're talking about coefficients. 00:57:09.360 --> 00:57:10.950 align:middle line:90% What are these coefficients? 00:57:10.950 --> 00:57:12.690 align:middle line:90% What are the coefficients of? 00:57:12.690 --> 00:57:15.590 align:middle line:84% Where are we getting the coefficients? 00:57:15.590 --> 00:57:17.000 align:middle line:90% AUDIENCE: Of x or y. 00:57:17.000 --> 00:57:18.625 align:middle line:84% BRYNMOR CHAPMAN: Not quite of x and y. 00:57:18.625 --> 00:57:25.700 align:middle line:90% 00:57:25.700 --> 00:57:29.016 align:middle line:84% So we're going to want to group like terms. 00:57:29.016 --> 00:57:30.950 align:middle line:84% So both multiplication and addition 00:57:30.950 --> 00:57:34.780 align:middle line:84% are commutative, so we can fiddle around 00:57:34.780 --> 00:57:37.840 align:middle line:84% with each of those terms and put all the x's at the beginning 00:57:37.840 --> 00:57:39.250 align:middle line:90% and the y's at the end. 00:57:39.250 --> 00:57:41.240 align:middle line:84% And then we can also rearrange the terms 00:57:41.240 --> 00:57:44.260 align:middle line:84% so that we put like terms together. 00:57:44.260 --> 00:57:56.380 align:middle line:84% So we can group them into a bunch of terms of the form x 00:57:56.380 --> 00:58:01.390 align:middle line:84% to the k times y to the n minus k 00:58:01.390 --> 00:58:04.380 align:middle line:84% and then multiplied by some coefficient c sub k. 00:58:04.380 --> 00:58:11.930 align:middle line:90% 00:58:11.930 --> 00:58:14.830 align:middle line:84% So now, as your colleague was saying, 00:58:14.830 --> 00:58:17.650 align:middle line:84% those c sub k are going to be the entries 00:58:17.650 --> 00:58:18.620 align:middle line:90% of Pascal's triangle. 00:58:18.620 --> 00:58:21.560 align:middle line:84% They're going to be the binomial coefficients. 00:58:21.560 --> 00:58:22.155 align:middle line:90% Why is that? 00:58:22.155 --> 00:58:26.200 align:middle line:90% 00:58:26.200 --> 00:58:29.370 align:middle line:84% Well, how many terms are there of this form? 00:58:29.370 --> 00:58:36.300 align:middle line:90% 00:58:36.300 --> 00:58:42.450 align:middle line:84% Well, it could come from putting the kx's in any k places 00:58:42.450 --> 00:58:46.410 align:middle line:84% from among the n possible places. 00:58:46.410 --> 00:58:49.380 align:middle line:84% So you've got n choose k possible places or possible ways 00:58:49.380 --> 00:58:51.030 align:middle line:90% to place the x's. 00:58:51.030 --> 00:58:54.815 align:middle line:84% The y's will be the remaining ones, and each of those 00:58:54.815 --> 00:58:55.940 align:middle line:90% gives you a different term. 00:58:55.940 --> 00:58:59.784 align:middle line:90% 00:58:59.784 --> 00:59:06.830 align:middle line:84% So c sub k is going to be n choose k, similar to above. 00:59:06.830 --> 00:59:12.420 align:middle line:90% 00:59:12.420 --> 00:59:14.200 align:middle line:84% Does that make sense to everybody? 00:59:14.200 --> 00:59:20.340 align:middle line:90% 00:59:20.340 --> 00:59:23.670 align:middle line:84% So if we specifically put in x and y are both equal to 1, 00:59:23.670 --> 00:59:26.500 align:middle line:84% then we recover what we had originally. 00:59:26.500 --> 00:59:30.380 align:middle line:84% You may also have encountered a form of this theorem 00:59:30.380 --> 00:59:32.540 align:middle line:90% where y equals 1. 00:59:32.540 --> 00:59:36.130 align:middle line:84% Have people seen that before, like 1 plus x to the n? 00:59:36.130 --> 00:59:41.000 align:middle line:90% 00:59:41.000 --> 00:59:41.510 align:middle line:90% Possibly? 00:59:41.510 --> 00:59:42.130 align:middle line:90% Maybe not? 00:59:42.130 --> 00:59:46.520 align:middle line:90% 00:59:46.520 --> 00:59:49.290 align:middle line:90% Well, it is a useful theorem. 00:59:49.290 --> 00:59:52.820 align:middle line:84% The coefficients on the terms are the n choose k, 00:59:52.820 --> 00:59:55.322 align:middle line:84% the binomial coefficients, and this 00:59:55.322 --> 00:59:56.530 align:middle line:90% is where the name comes from. 00:59:56.530 --> 00:59:59.474 align:middle line:90% 00:59:59.474 --> 01:00:02.750 align:middle line:84% And more generally, you can also get the multinomial theorem 01:00:02.750 --> 01:00:04.692 align:middle line:90% in the same way. 01:00:04.692 --> 01:00:09.210 align:middle line:84% If you have x plus y plus z for instance, 01:00:09.210 --> 01:00:10.897 align:middle line:90% raise that to the n-th power. 01:00:10.897 --> 01:00:12.980 align:middle line:84% You start getting multinomial coefficients instead 01:00:12.980 --> 01:00:14.550 align:middle line:90% of binomial coefficients. 01:00:14.550 --> 01:00:50.040 align:middle line:90% 01:00:50.040 --> 01:00:50.540 align:middle line:90% Yeah. 01:00:50.540 --> 01:00:52.130 align:middle line:84% Sorry, I'm moving all the way over here. 01:00:52.130 --> 01:00:53.550 align:middle line:84% I didn't plan this out very well. 01:00:53.550 --> 01:01:18.460 align:middle line:90% 01:01:18.460 --> 01:01:22.340 align:middle line:84% So for the multinomial theorem, if we have a multinomial here, 01:01:22.340 --> 01:01:26.926 align:middle line:84% like we've got mx's instead of just two, 01:01:26.926 --> 01:01:29.900 align:middle line:84% if we sum all those up and then raise it to the n-th power, 01:01:29.900 --> 01:01:32.620 align:middle line:84% we can use essentially the same proof 01:01:32.620 --> 01:01:39.430 align:middle line:84% to show that this is going to be the sum over all sequences k1, 01:01:39.430 --> 01:01:46.240 align:middle line:84% k2, all the way up to k sub m, where these all sum up 01:01:46.240 --> 01:01:56.260 align:middle line:84% to n of the multinomial coefficient n choose k1, k2, 01:01:56.260 --> 01:02:03.790 align:middle line:84% up to km of similar to the x to the ky to the n minus k 01:02:03.790 --> 01:02:06.130 align:middle line:90% term we had over there. 01:02:06.130 --> 01:02:08.153 align:middle line:84% This time, we've got a product of m things. 01:02:08.153 --> 01:02:17.110 align:middle line:90% 01:02:17.110 --> 01:02:21.400 align:middle line:90% So essentially the same proof. 01:02:21.400 --> 01:02:27.210 align:middle line:84% If we think of the binomial coefficients 01:02:27.210 --> 01:02:32.370 align:middle line:84% as multinomial coefficients where we've only got two things, 01:02:32.370 --> 01:02:33.740 align:middle line:90% you can see how it expands. 01:02:33.740 --> 01:02:38.570 align:middle line:90% 01:02:38.570 --> 01:02:41.460 align:middle line:84% We're going to look at all possible ways 01:02:41.460 --> 01:02:47.820 align:middle line:84% that you could place x1's, x2's, x3's, et cetera 01:02:47.820 --> 01:02:55.260 align:middle line:84% among the n different terms, and each of those 01:02:55.260 --> 01:02:59.490 align:middle line:84% is going to give us a different term that we can then combine 01:02:59.490 --> 01:03:01.270 align:middle line:90% when we're grouping like terms. 01:03:01.270 --> 01:03:04.830 align:middle line:90% Does that make sense to people? 01:03:04.830 --> 01:03:06.570 align:middle line:84% So the number of those, the number 01:03:06.570 --> 01:03:11.430 align:middle line:84% of ways that you can permute those 01:03:11.430 --> 01:03:15.735 align:middle line:84% n things, that's just going to be this multinomial coefficient 01:03:15.735 --> 01:03:16.235 align:middle line:90% here. 01:03:16.235 --> 01:03:43.880 align:middle line:90% 01:03:43.880 --> 01:03:46.370 align:middle line:84% Are people happy with the binomial multinomial 01:03:46.370 --> 01:03:49.920 align:middle line:84% coefficients and the binomial multinomial theorems? 01:03:49.920 --> 01:03:50.780 align:middle line:90% Yeah. 01:03:50.780 --> 01:03:53.580 align:middle line:84% AUDIENCE: So the coefficients, what is that? 01:03:53.580 --> 01:04:01.130 align:middle line:84% Like n over k1 comma k2 comma [INAUDIBLE]? 01:04:01.130 --> 01:04:02.030 align:middle line:90% BRYNMOR CHAPMAN: So-- 01:04:02.030 --> 01:04:02.960 align:middle line:90% AUDIENCE: [INAUDIBLE] 01:04:02.960 --> 01:04:06.860 align:middle line:84% BRYNMOR CHAPMAN: The question is, what does this mean? 01:04:06.860 --> 01:04:08.360 align:middle line:84% Is this the notation that you saw 01:04:08.360 --> 01:04:12.080 align:middle line:90% in the bookkeeper recitation? 01:04:12.080 --> 01:04:12.810 align:middle line:90% Perhaps not. 01:04:12.810 --> 01:04:17.600 align:middle line:84% If not, this is equal to n factorial divided 01:04:17.600 --> 01:04:25.340 align:middle line:84% by k1 factorial, k2 factorial, all the way up to km factorial. 01:04:25.340 --> 01:04:28.010 align:middle line:90% 01:04:28.010 --> 01:04:31.510 align:middle line:84% So it's kind of an extension of the binomial coefficients. 01:04:31.510 --> 01:04:33.370 align:middle line:84% With the binomial coefficients, you just 01:04:33.370 --> 01:04:37.510 align:middle line:90% have two k's that sum to n. 01:04:37.510 --> 01:04:39.310 align:middle line:84% And we also simplify the notation 01:04:39.310 --> 01:04:41.680 align:middle line:90% by just dropping one of the-- 01:04:41.680 --> 01:04:45.730 align:middle line:84% dropping one of the k's because it's redundant in that case. 01:04:45.730 --> 01:04:47.860 align:middle line:84% But for the multinomial coefficients, 01:04:47.860 --> 01:04:52.360 align:middle line:84% we have a bunch of k's that sum to n instead of just two. 01:04:52.360 --> 01:04:53.800 align:middle line:84% We're going to divide n factorial 01:04:53.800 --> 01:04:57.525 align:middle line:90% by the product of all of them. 01:04:57.525 --> 01:04:58.400 align:middle line:90% Does that make sense? 01:04:58.400 --> 01:05:01.651 align:middle line:90% Does that answer your question? 01:05:01.651 --> 01:05:05.290 align:middle line:84% Any other questions about binomial multinomial theorems? 01:05:05.290 --> 01:05:07.480 align:middle line:90% Yeah. 01:05:07.480 --> 01:05:08.240 align:middle line:90% Sorry? 01:05:08.240 --> 01:05:09.115 align:middle line:90% AUDIENCE: [INAUDIBLE] 01:05:09.115 --> 01:05:11.140 align:middle line:84% BRYNMOR CHAPMAN: Oh, question-- what is the pi? 01:05:11.140 --> 01:05:13.930 align:middle line:84% So this is an analog of sigma but for products 01:05:13.930 --> 01:05:15.310 align:middle line:90% instead of sums. 01:05:15.310 --> 01:05:19.920 align:middle line:84% So we're saying xi to the ki multiplied 01:05:19.920 --> 01:05:27.450 align:middle line:84% over all possible i, so x1 to the k1 times x2 to the k2 times 01:05:27.450 --> 01:05:29.573 align:middle line:90% x3 to the k3, cetera. 01:05:29.573 --> 01:05:33.828 align:middle line:90% 01:05:33.828 --> 01:05:35.690 align:middle line:90% So pi for product. 01:05:35.690 --> 01:05:43.890 align:middle line:90% 01:05:43.890 --> 01:05:45.116 align:middle line:90% Any other questions? 01:05:45.116 --> 01:05:51.051 align:middle line:90% 01:05:51.051 --> 01:05:53.460 align:middle line:84% So why don't we move on to another useful identity 01:05:53.460 --> 01:05:56.120 align:middle line:84% that we can prove using a combinatorial proof? 01:05:56.120 --> 01:06:00.870 align:middle line:90% 01:06:00.870 --> 01:06:07.470 align:middle line:84% n choose k equals n minus 1 choose k minus 1 01:06:07.470 --> 01:06:11.375 align:middle line:90% plus n minus 1 choose k. 01:06:11.375 --> 01:06:16.530 align:middle line:90% 01:06:16.530 --> 01:06:17.710 align:middle line:90% Who has seen this before? 01:06:17.710 --> 01:06:19.210 align:middle line:84% I know that at least one of you has. 01:06:19.210 --> 01:06:24.150 align:middle line:90% 01:06:24.150 --> 01:06:24.845 align:middle line:90% Few hands. 01:06:24.845 --> 01:06:28.590 align:middle line:90% 01:06:28.590 --> 01:06:31.720 align:middle line:84% So we could prove it algebraically, 01:06:31.720 --> 01:06:33.870 align:middle line:84% just writing them out with factorials 01:06:33.870 --> 01:06:37.800 align:middle line:84% and doing a bunch of fiddly garbage, 01:06:37.800 --> 01:06:40.560 align:middle line:84% but that seems a bit difficult. So let's use a double counting 01:06:40.560 --> 01:06:41.265 align:middle line:90% proof instead. 01:06:41.265 --> 01:06:44.040 align:middle line:90% 01:06:44.040 --> 01:06:46.140 align:middle line:84% How might we use a double counting proof 01:06:46.140 --> 01:06:47.830 align:middle line:90% to prove this identity? 01:06:47.830 --> 01:06:53.730 align:middle line:90% 01:06:53.730 --> 01:06:55.715 align:middle line:90% What set could we try and count? 01:06:55.715 --> 01:07:01.470 align:middle line:90% 01:07:01.470 --> 01:07:02.345 align:middle line:90% Any ideas? 01:07:02.345 --> 01:07:08.130 align:middle line:90% 01:07:08.130 --> 01:07:09.884 align:middle line:90% Yeah. 01:07:09.884 --> 01:07:12.330 align:middle line:84% AUDIENCE: [INAUDIBLE] once again, consider 01:07:12.330 --> 01:07:15.900 align:middle line:84% the binary sequence [INAUDIBLE] that has k1's in it 01:07:15.900 --> 01:07:16.960 align:middle line:90% and [INAUDIBLE]? 01:07:16.960 --> 01:07:20.410 align:middle line:90% 01:07:20.410 --> 01:07:22.100 align:middle line:84% BRYNMOR CHAPMAN: Yeah, good answer. 01:07:22.100 --> 01:07:26.170 align:middle line:84% So the answer was, once again, we can look at binary sequences 01:07:26.170 --> 01:07:29.350 align:middle line:84% of length n, but this time, we're going to restrict it 01:07:29.350 --> 01:07:33.403 align:middle line:84% to binary sequences of length n that specifically have k1's. 01:07:33.403 --> 01:07:35.920 align:middle line:90% 01:07:35.920 --> 01:07:51.563 align:middle line:84% So count binary sequences of length n with k1's. 01:07:51.563 --> 01:07:58.663 align:middle line:90% 01:07:58.663 --> 01:08:00.225 align:middle line:84% Now, how many of those are there? 01:08:00.225 --> 01:08:05.140 align:middle line:90% 01:08:05.140 --> 01:08:07.755 align:middle line:84% Does that give us one of our two expressions easily? 01:08:07.755 --> 01:08:11.380 align:middle line:90% 01:08:11.380 --> 01:08:13.960 align:middle line:90% If so, which? 01:08:13.960 --> 01:08:14.755 align:middle line:90% Yeah? 01:08:14.755 --> 01:08:15.630 align:middle line:90% AUDIENCE: [INAUDIBLE] 01:08:15.630 --> 01:08:16.775 align:middle line:90% BRYNMOR CHAPMAN: Sorry? 01:08:16.775 --> 01:08:17.040 align:middle line:90% AUDIENCE: [INAUDIBLE] 01:08:17.040 --> 01:08:18.060 align:middle line:90% BRYNMOR CHAPMAN: Yeah. 01:08:18.060 --> 01:08:20.729 align:middle line:84% So essentially by definition, like this 01:08:20.729 --> 01:08:23.614 align:middle line:84% is equal to-- the size of this set is going to be n choose k. 01:08:23.614 --> 01:08:32.080 align:middle line:90% 01:08:32.080 --> 01:08:48.029 align:middle line:90% So let's call this set S. 01:08:48.029 --> 01:08:51.500 align:middle line:84% How could we also make S equal to the right-hand side? 01:08:51.500 --> 01:09:19.430 align:middle line:90% 01:09:19.430 --> 01:09:20.724 align:middle line:90% Does this give us any ideas? 01:09:20.724 --> 01:09:30.830 align:middle line:90% 01:09:30.830 --> 01:09:32.000 align:middle line:90% Yeah, answer? 01:09:32.000 --> 01:09:40.713 align:middle line:84% AUDIENCE: [INAUDIBLE] it can be either 0 or 1. 01:09:40.713 --> 01:09:41.630 align:middle line:90% BRYNMOR CHAPMAN: Yeah. 01:09:41.630 --> 01:09:46.069 align:middle line:84% AUDIENCE: [INAUDIBLE] you want to have k1 and then pick 01:09:46.069 --> 01:09:48.560 align:middle line:84% an option [INAUDIBLE] original [INAUDIBLE]. 01:09:48.560 --> 01:09:51.843 align:middle line:90% 01:09:51.843 --> 01:09:52.760 align:middle line:90% BRYNMOR CHAPMAN: Yeah. 01:09:52.760 --> 01:09:54.650 align:middle line:84% So the answer was, essentially, we're 01:09:54.650 --> 01:09:58.166 align:middle line:90% going to partition our set. 01:09:58.166 --> 01:10:00.840 align:middle line:90% So what are we partitioning on? 01:10:00.840 --> 01:10:03.670 align:middle line:84% Well, we're going to look at the last bit of the sequence. 01:10:03.670 --> 01:10:25.160 align:middle line:90% 01:10:25.160 --> 01:10:26.990 align:middle line:90% So we've got two cases. 01:10:26.990 --> 01:10:29.740 align:middle line:84% The last bit can either be 0 or 1. 01:10:29.740 --> 01:10:32.320 align:middle line:90% It can't be both. 01:10:32.320 --> 01:10:37.900 align:middle line:90% So we've got a partition. 01:10:37.900 --> 01:10:40.660 align:middle line:84% That means that we can use the sum rule. 01:10:40.660 --> 01:10:45.610 align:middle line:84% We can count each of these sets individually and then 01:10:45.610 --> 01:10:46.922 align:middle line:90% add their sizes together. 01:10:46.922 --> 01:10:48.880 align:middle line:84% That will give us the size of our original set. 01:10:48.880 --> 01:10:51.910 align:middle line:90% 01:10:51.910 --> 01:10:58.410 align:middle line:84% So how many length n sequences with k1's have a 1 at the end? 01:10:58.410 --> 01:11:08.860 align:middle line:90% 01:11:08.860 --> 01:11:09.430 align:middle line:90% Yeah? 01:11:09.430 --> 01:11:11.880 align:middle line:90% AUDIENCE: [INAUDIBLE] 01:11:11.880 --> 01:11:13.680 align:middle line:90% BRYNMOR CHAPMAN: Yeah. 01:11:13.680 --> 01:11:18.005 align:middle line:84% So the answer was n minus 1 choose k minus 1. 01:11:18.005 --> 01:11:45.006 align:middle line:90% 01:11:45.006 --> 01:11:52.423 align:middle line:84% So let's partition this S, S0 union S1. 01:11:52.423 --> 01:11:56.910 align:middle line:90% 01:11:56.910 --> 01:12:00.480 align:middle line:90% S0-- or let's call it S sub b-- 01:12:00.480 --> 01:12:11.570 align:middle line:84% is the set of binary sequences with n1's-- 01:12:11.570 --> 01:12:20.125 align:middle line:84% sorry-- n bits, k1's, and b at the end. 01:12:20.125 --> 01:12:24.950 align:middle line:90% 01:12:24.950 --> 01:12:35.300 align:middle line:84% So S sub 1 is going to have size n minus 1 choose k minus 1. 01:12:35.300 --> 01:12:38.280 align:middle line:84% If we know that we've got a 1 at the end, 01:12:38.280 --> 01:12:41.900 align:middle line:84% we've got k minus 1 additional 1's that we need to place. 01:12:41.900 --> 01:12:43.670 align:middle line:90% Where can they go? 01:12:43.670 --> 01:12:46.846 align:middle line:84% Into any of the first n minus 1 positions. 01:12:46.846 --> 01:12:50.990 align:middle line:90% 01:12:50.990 --> 01:12:56.330 align:middle line:84% So there are n minus 1 choose k minus 1 ways to choose those. 01:12:56.330 --> 01:12:57.610 align:middle line:90% And what about S0? 01:12:57.610 --> 01:13:13.360 align:middle line:90% 01:13:13.360 --> 01:13:14.170 align:middle line:90% Yeah. 01:13:14.170 --> 01:13:15.580 align:middle line:90% AUDIENCE: n minus 1 choose k. 01:13:15.580 --> 01:13:17.770 align:middle line:90% BRYNMOR CHAPMAN: Yeah. 01:13:17.770 --> 01:13:20.230 align:middle line:84% This is going to be n minus 1 choose k. 01:13:20.230 --> 01:13:22.562 align:middle line:90% And why is that? 01:13:22.562 --> 01:13:24.473 align:middle line:84% AUDIENCE: Because there are still k1's. 01:13:24.473 --> 01:13:25.390 align:middle line:90% BRYNMOR CHAPMAN: Yeah. 01:13:25.390 --> 01:13:28.600 align:middle line:84% If we've got a 0 at the end, we still 01:13:28.600 --> 01:13:31.210 align:middle line:84% need to place those k1's somewhere. 01:13:31.210 --> 01:13:35.120 align:middle line:84% So now we've only got n minus 1 places where we can put them, 01:13:35.120 --> 01:13:37.360 align:middle line:84% but we still have k1's that we need to place. 01:13:37.360 --> 01:13:40.050 align:middle line:84% So there are n minus 1 choose k ways to place them. 01:13:40.050 --> 01:13:48.975 align:middle line:90% 01:13:48.975 --> 01:13:50.990 align:middle line:84% So these two cases are exhaustive. 01:13:50.990 --> 01:13:54.580 align:middle line:84% You can't have-- you can't have a sequence that 01:13:54.580 --> 01:13:56.240 align:middle line:90% ends with neither a 0 or a 1. 01:13:56.240 --> 01:13:58.310 align:middle line:90% It's a binary sequence. 01:13:58.310 --> 01:13:59.240 align:middle line:90% They're also disjoint. 01:13:59.240 --> 01:14:01.780 align:middle line:84% You can't have both a 0 and a 1 at the end. 01:14:01.780 --> 01:14:04.370 align:middle line:90% So this forms a partition. 01:14:04.370 --> 01:14:11.310 align:middle line:84% So by the sum rule, the size of S is going to be the size of S0 01:14:11.310 --> 01:14:17.533 align:middle line:84% plus the size of S1, which, we just established, is-- 01:14:17.533 --> 01:14:18.825 align:middle line:90% I guess I flipped them around-- 01:14:18.825 --> 01:14:22.352 align:middle line:90% 01:14:22.352 --> 01:14:26.690 align:middle line:84% n minus 1 choose k plus n minus 1 choose k minus 1. 01:14:26.690 --> 01:14:31.260 align:middle line:90% 01:14:31.260 --> 01:14:33.600 align:middle line:84% And as we established at the beginning, 01:14:33.600 --> 01:14:37.145 align:middle line:84% this is also equal to n choose k. 01:14:37.145 --> 01:14:40.780 align:middle line:90% 01:14:40.780 --> 01:14:42.730 align:middle line:84% Does that make sense to everybody? 01:14:42.730 --> 01:15:03.330 align:middle line:90% 01:15:03.330 --> 01:15:06.910 align:middle line:84% Do people have any questions about either this identity 01:15:06.910 --> 01:15:09.970 align:middle line:90% or combinatorial proofs? 01:15:09.970 --> 01:15:10.608 align:middle line:90% Yeah. 01:15:10.608 --> 01:15:13.150 align:middle line:84% AUDIENCE: So the reason why we didn't just write [INAUDIBLE]? 01:15:13.150 --> 01:15:18.743 align:middle line:90% 01:15:18.743 --> 01:15:19.660 align:middle line:90% BRYNMOR CHAPMAN: Yeah. 01:15:19.660 --> 01:15:21.220 align:middle line:84% The question was, is there any reason 01:15:21.220 --> 01:15:24.350 align:middle line:84% we didn't write out the definition of n 01:15:24.350 --> 01:15:28.390 align:middle line:84% choose k using the factorial, simplify things? 01:15:28.390 --> 01:15:31.962 align:middle line:84% Because I'm a lazy fuck, and I don't want to. 01:15:31.962 --> 01:15:36.593 align:middle line:84% AUDIENCE: [INAUDIBLE] a lot longer than that [INAUDIBLE]. 01:15:36.593 --> 01:15:37.510 align:middle line:90% BRYNMOR CHAPMAN: Yeah. 01:15:37.510 --> 01:15:41.090 align:middle line:84% So that kind of depends on-- so the concern 01:15:41.090 --> 01:15:46.747 align:middle line:84% was that maybe this takes a lot longer than just 01:15:46.747 --> 01:15:48.580 align:middle line:84% putting your head down and going through all 01:15:48.580 --> 01:15:51.440 align:middle line:90% of the messy algebra. 01:15:51.440 --> 01:15:51.940 align:middle line:90% Yeah. 01:15:51.940 --> 01:15:54.772 align:middle line:84% At the beginning, it probably will. 01:15:54.772 --> 01:15:58.040 align:middle line:84% If you are not familiar with this proof technique, 01:15:58.040 --> 01:16:01.210 align:middle line:84% it may not be obvious what set you should be counting 01:16:01.210 --> 01:16:06.410 align:middle line:84% or how you should be counting it, but as you get practice, 01:16:06.410 --> 01:16:10.650 align:middle line:84% it could be much easier than just trying to brute-force it. 01:16:10.650 --> 01:16:16.830 align:middle line:84% And I think in possibly in recitation, in homework, 01:16:16.830 --> 01:16:20.300 align:middle line:84% you will see other identities where 01:16:20.300 --> 01:16:24.127 align:middle line:84% it may be less viable to actually go 01:16:24.127 --> 01:16:25.210 align:middle line:90% through the messy algebra. 01:16:25.210 --> 01:16:29.099 align:middle line:90% 01:16:29.099 --> 01:16:31.395 align:middle line:84% So yeah, this is kind of a silly example, 01:16:31.395 --> 01:16:33.020 align:middle line:84% and it's one where you could definitely 01:16:33.020 --> 01:16:35.540 align:middle line:90% solve it in other ways. 01:16:35.540 --> 01:16:38.580 align:middle line:84% And in fact, those of you who have seen this identity before, 01:16:38.580 --> 01:16:42.876 align:middle line:84% I imagine you probably did prove it in other ways. 01:16:42.876 --> 01:16:44.960 align:middle line:84% But it's just kind of illustrative 01:16:44.960 --> 01:16:48.080 align:middle line:84% of this particular proof technique. 01:16:48.080 --> 01:16:49.470 align:middle line:90% Does that answer your question? 01:16:49.470 --> 01:16:50.210 align:middle line:90% AUDIENCE: Yeah. 01:16:50.210 --> 01:16:52.710 align:middle line:90% BRYNMOR CHAPMAN: OK. 01:16:52.710 --> 01:16:54.540 align:middle line:84% Any other questions, comments, concerns? 01:16:54.540 --> 01:17:12.080 align:middle line:90% 01:17:12.080 --> 01:17:21.610 align:middle line:84% So as your colleague was hinting at earlier, 01:17:21.610 --> 01:17:25.450 align:middle line:84% this kind of gives us the following. 01:17:25.450 --> 01:17:28.930 align:middle line:84% So I imagine many of you have seen Pascal's triangle before, 01:17:28.930 --> 01:17:35.710 align:middle line:84% but if you just write out all of the binomial coefficients 01:17:35.710 --> 01:17:38.510 align:middle line:90% in rows of a triangle-- 01:17:38.510 --> 01:17:43.840 align:middle line:90% 01:17:43.840 --> 01:17:52.100 align:middle line:84% so this is n equals 0, n equals 1, n equals 2, n equals 3. 01:17:52.100 --> 01:17:55.540 align:middle line:84% And then successive entries in a row 01:17:55.540 --> 01:17:59.230 align:middle line:84% are going to be n choose 0, n choose 1, n choose 2, et cetera. 01:17:59.230 --> 01:18:13.350 align:middle line:84% So if you write all of these out, what we just proved 01:18:13.350 --> 01:18:16.590 align:middle line:84% is that every entry of this triangle 01:18:16.590 --> 01:18:22.320 align:middle line:84% is the sum of the two entries above it. 01:18:22.320 --> 01:18:28.110 align:middle line:84% So if we take this 4 and the 6 here, they sum to 10. 01:18:28.110 --> 01:18:32.535 align:middle line:84% That is essentially what this identity is saying. 01:18:32.535 --> 01:18:35.640 align:middle line:90% 01:18:35.640 --> 01:18:40.500 align:middle line:84% So this is what we call Pascal's triangle. 01:18:40.500 --> 01:18:43.980 align:middle line:84% We also showed that the sum of all 01:18:43.980 --> 01:18:48.340 align:middle line:84% of the numbers in a single row is going to be a power of 2. 01:18:48.340 --> 01:18:50.660 align:middle line:90% It's going to be 2 to the n. 01:18:50.660 --> 01:18:55.530 align:middle line:90% 01:18:55.530 --> 01:19:00.950 align:middle line:84% If we sum everything in row 5, we get 1 plus 5 plus 10 01:19:00.950 --> 01:19:04.870 align:middle line:90% plus 10 plus 5 plus 1 equals 32. 01:19:04.870 --> 01:19:10.040 align:middle line:90% 01:19:10.040 --> 01:19:11.730 align:middle line:90% I think we proved it over there. 01:19:11.730 --> 01:19:14.500 align:middle line:84% Yeah, top of the rightmost board. 01:19:14.500 --> 01:19:17.710 align:middle line:90% 01:19:17.710 --> 01:19:24.710 align:middle line:84% And you will see, I believe, in homework 01:19:24.710 --> 01:19:28.860 align:middle line:84% that if you sum the diagonals, you get the Fibonacci numbers. 01:19:28.860 --> 01:19:30.956 align:middle line:90% So if you kind of-- 01:19:30.956 --> 01:19:32.720 align:middle line:90% oh, let's see. 01:19:32.720 --> 01:19:35.323 align:middle line:90% So go like that. 01:19:35.323 --> 01:19:41.141 align:middle line:90% 01:19:41.141 --> 01:19:42.650 align:middle line:84% If you sum along these diagonals, 01:19:42.650 --> 01:19:45.230 align:middle line:90% you get the Fibonacci numbers. 01:19:45.230 --> 01:19:45.840 align:middle line:90% Am I correct? 01:19:45.840 --> 01:19:47.215 align:middle line:84% Is that still a homework problem? 01:19:47.215 --> 01:19:49.550 align:middle line:90% 01:19:49.550 --> 01:19:50.300 align:middle line:90% Maybe? 01:19:50.300 --> 01:19:52.430 align:middle line:84% Sorry, I've not actually looked at your homework. 01:19:52.430 --> 01:19:53.370 align:middle line:90% Full disclosure. 01:19:53.370 --> 01:19:56.510 align:middle line:90% 01:19:56.510 --> 01:20:00.640 align:middle line:84% It's a fun exercise if you want to try it. 01:20:00.640 --> 01:20:05.350 align:middle line:84% But I will leave you with one more fun exercise. 01:20:05.350 --> 01:20:14.980 align:middle line:84% Prove that the sum from k equals 0 to n n choose k squared is 01:20:14.980 --> 01:20:17.560 align:middle line:90% going to be 2n choose n. 01:20:17.560 --> 01:20:20.590 align:middle line:90% 01:20:20.590 --> 01:20:24.100 align:middle line:84% So if you look at all of the entries of a row, 01:20:24.100 --> 01:20:26.150 align:middle line:84% instead of just summing them up directly, 01:20:26.150 --> 01:20:27.830 align:middle line:90% square them and sum them up. 01:20:27.830 --> 01:20:34.300 align:middle line:84% You get the entry that's twice as far down in the middle. 01:20:34.300 --> 01:20:39.580 align:middle line:84% So see if you can prove that using a double counting proof. 01:20:39.580 --> 01:20:41.710 align:middle line:84% So that's all we have time for today. 01:20:41.710 --> 01:20:44.600 align:middle line:84% If you have any questions, comments, concerns, 01:20:44.600 --> 01:20:47.590 align:middle line:90% please feel free to come up. 01:20:47.590 --> 01:20:50.350 align:middle line:84% Otherwise, good luck on the exam. 01:20:50.350 --> 01:20:51.900 align:middle line:90% Yeah. 01:20:51.900 --> 01:20:57.000 align:middle line:90%