WEBVTT 00:00:00.590 --> 00:00:05.150 In this segment we introduce a simple but powerful tool, the 00:00:05.150 --> 00:00:08.450 basic counting principle, which we will be using over 00:00:08.450 --> 00:00:11.820 and over to deal with counting problems. 00:00:11.820 --> 00:00:15.070 Let me describe the idea through a simple example. 00:00:15.070 --> 00:00:18.380 You wake up in the morning and you find that you have in your 00:00:18.380 --> 00:00:23.120 closet 4 shirts, 3 ties, and 2 jackets. 00:00:23.120 --> 00:00:27.000 In how many different ways can you get dressed today? 00:00:27.000 --> 00:00:30.580 To answer this question, let us think of the process of 00:00:30.580 --> 00:00:36.220 getting dressed as consisting of three steps, three stages. 00:00:36.220 --> 00:00:40.910 You first choose a shirt, let's say this one, and you 00:00:40.910 --> 00:00:44.600 have 4 choices of shirts. 00:00:44.600 --> 00:00:50.600 But each shirt can be used together with 1 of the 3 00:00:50.600 --> 00:00:55.710 available ties to make 3 different shirt-tie 00:00:55.710 --> 00:00:58.520 combinations. 00:00:58.520 --> 00:01:03.540 But since we had 4 choices for the shirt, this means that we 00:01:03.540 --> 00:01:09.970 have 4 times 3, equals 12, shirt-tie combinations. 00:01:09.970 --> 00:01:12.160 Finally, you choose a jacket. 00:01:12.160 --> 00:01:15.240 Each shirt-tie combination can go together with either 00:01:15.240 --> 00:01:19.400 jacket, and so the fact that you have 2 jackets available 00:01:19.400 --> 00:01:25.320 doubles the number of options that you have, leading to 24 00:01:25.320 --> 00:01:27.340 different options overall. 00:01:27.340 --> 00:01:31.190 So 24 is the answer to this simple problem. 00:01:31.190 --> 00:01:33.530 And how did the number 24 come about? 00:01:33.530 --> 00:01:37.420 Well, 24 is the same as the number of options you had in 00:01:37.420 --> 00:01:40.009 the first stage times the number of options you had in 00:01:40.009 --> 00:01:43.530 the second stage times the number of options you had in 00:01:43.530 --> 00:01:45.539 the third stage. 00:01:45.539 --> 00:01:48.610 Let us generalize. 00:01:48.610 --> 00:01:51.840 Suppose we want to construct some kind of object, and we're 00:01:51.840 --> 00:01:55.110 going to construct it through a sequential process, through 00:01:55.110 --> 00:01:58.890 a sequence of r different stages. 00:01:58.890 --> 00:02:02.140 In the example that we just considered, the number of 00:02:02.140 --> 00:02:05.600 stages was equal to 3. 00:02:05.600 --> 00:02:09.600 At each one of the stages, you have a number of options that 00:02:09.600 --> 00:02:10.699 are available. 00:02:10.699 --> 00:02:14.500 So in our example, at the first stage we had 4 options, 00:02:14.500 --> 00:02:19.270 at the second stage we had 3 options, and at the last stage 00:02:19.270 --> 00:02:21.990 we had 2 options. 00:02:21.990 --> 00:02:27.930 What is important is that when you reach stage i, no matter 00:02:27.930 --> 00:02:31.430 what you chose, no matter what you did at the previous 00:02:31.430 --> 00:02:35.329 stages, the number of options that you will have available 00:02:35.329 --> 00:02:42.180 at stage i is going to be that fixed number, n-sub-i. 00:02:42.180 --> 00:02:44.200 So what is the answer? 00:02:44.200 --> 00:02:47.500 How many different objects can you construct this way? 00:02:47.500 --> 00:02:50.950 Well, just generalizing from what we did in our specific 00:02:50.950 --> 00:02:56.500 example, the answer is the product of the number of 00:02:56.500 --> 00:03:02.950 choices or options that you had at each stage. 00:03:02.950 --> 00:03:05.240 This is the counting principle. 00:03:05.240 --> 00:03:08.740 It's a very simple idea, but it is powerful. 00:03:08.740 --> 00:03:11.140 It will allow us to solve fairly 00:03:11.140 --> 00:03:13.790 complicated counting problems. 00:03:13.790 --> 00:03:17.990 However, before we go into more complicated problems, let 00:03:17.990 --> 00:03:21.760 us first deal with a few relatively easy examples. 00:03:26.300 --> 00:03:31.660 In our first example, let us consider license plates that 00:03:31.660 --> 00:03:35.240 consist of 2 letters followed by 3 digits. 00:03:35.240 --> 00:03:37.320 The question is, how many different 00:03:37.320 --> 00:03:39.620 license plates are there? 00:03:39.620 --> 00:03:43.250 We think of the process of constructing a license plate 00:03:43.250 --> 00:03:45.130 as a sequential process. 00:03:45.130 --> 00:03:51.700 At the first stage we choose a letter, and we have 26 choices 00:03:51.700 --> 00:03:53.460 for the first letter. 00:03:53.460 --> 00:03:57.690 Then we need to choose the second letter, and we have 26 00:03:57.690 --> 00:03:59.700 choices for that one. 00:03:59.700 --> 00:04:01.210 Then we choose the first digit. 00:04:01.210 --> 00:04:03.160 We have 10 choices for it. 00:04:03.160 --> 00:04:06.870 We choose the second digit, for which we have 10 choices. 00:04:06.870 --> 00:04:09.460 And finally, we choose the last digit, for which we also 00:04:09.460 --> 00:04:10.790 have 10 choices. 00:04:10.790 --> 00:04:13.880 So if you multiply these numbers, you can find the 00:04:13.880 --> 00:04:16.560 number of different license plates that you can make with 00:04:16.560 --> 00:04:18.930 2 letters followed by 3 digits. 00:04:18.930 --> 00:04:23.540 Now let us change the problem a little bit and require that 00:04:23.540 --> 00:04:28.775 no letter and no digit can be used more than once. 00:04:31.300 --> 00:04:35.960 So, let us think of a process by which we could construct 00:04:35.960 --> 00:04:38.580 license plates of this kind. 00:04:38.580 --> 00:04:44.480 In the first stage, we choose the first letter that goes to 00:04:44.480 --> 00:04:48.450 the license plate, and we have 26 choices. 00:04:48.450 --> 00:04:51.620 Now, let us go into a second stage where we choose the 00:04:51.620 --> 00:04:54.070 second letter. 00:04:54.070 --> 00:04:57.930 Because we used 1 letter in the first stage, this means 00:04:57.930 --> 00:05:02.690 that there's only 25 available letters that can be used. 00:05:02.690 --> 00:05:06.900 We only have 25 choices at the second stage. 00:05:06.900 --> 00:05:08.790 Now, let us start dealing with the digits. 00:05:08.790 --> 00:05:12.530 We choose the first digit, and we have 10 choices for it. 00:05:12.530 --> 00:05:17.400 However, when we go and choose the next digit we will only 00:05:17.400 --> 00:05:20.780 have 9 choices, because 1 of the digits has 00:05:20.780 --> 00:05:22.620 already been used. 00:05:22.620 --> 00:05:26.450 At this point, 2 digits have been used, which means that at 00:05:26.450 --> 00:05:30.630 the last stage we have only 8 digits to choose from. 00:05:30.630 --> 00:05:34.590 So by multiplying these numbers, we can find out the 00:05:34.590 --> 00:05:37.590 answer to this question, the number of license plates if 00:05:37.590 --> 00:05:40.860 repetition is prohibited. 00:05:40.860 --> 00:05:43.830 Let us now consider a different example. 00:05:46.450 --> 00:05:49.450 Suppose that we start with a set that 00:05:49.450 --> 00:05:51.710 consists of n elements. 00:05:54.540 --> 00:05:57.130 What we want to do is to take these n 00:05:57.130 --> 00:06:00.230 elements and order them. 00:06:00.230 --> 00:06:03.650 A terminology that's often used here is that we want to 00:06:03.650 --> 00:06:07.290 form a permutation of these n elements. 00:06:07.290 --> 00:06:11.850 One way of visualizing permutations is to say that 00:06:11.850 --> 00:06:14.720 we're going to take these elements of the set, which are 00:06:14.720 --> 00:06:18.010 unordered, and we're going to place them 00:06:18.010 --> 00:06:20.590 in a sequence slots. 00:06:20.590 --> 00:06:23.030 So we create n slots. 00:06:26.390 --> 00:06:31.750 And we want to put each one of these elements into one of 00:06:31.750 --> 00:06:33.500 these slots. 00:06:33.500 --> 00:06:36.350 How do we go about it? 00:06:36.350 --> 00:06:39.760 We think of putting the elements into slots, 00:06:39.760 --> 00:06:41.890 one slot at a time. 00:06:41.890 --> 00:06:44.350 We first consider the first slot. 00:06:44.350 --> 00:06:48.740 We pick one of the elements and put it there. 00:06:48.740 --> 00:06:50.820 How many choices do we have at this stage? 00:06:50.820 --> 00:06:54.260 We have n choices, because we can pick any of the available 00:06:54.260 --> 00:06:56.930 elements and place it in that slot. 00:06:56.930 --> 00:07:00.420 Next, we pick another element and put it 00:07:00.420 --> 00:07:02.380 inside the second slot. 00:07:02.380 --> 00:07:04.765 How many choices do we have at this step? 00:07:04.765 --> 00:07:10.180 Well, we have already used one of the available elements, 00:07:10.180 --> 00:07:13.280 which means that there's n minus 1 elements to choose 00:07:13.280 --> 00:07:17.160 from at the next stage. 00:07:17.160 --> 00:07:20.190 At this point, we have used 2 of the elements. 00:07:20.190 --> 00:07:24.120 There is n minus 2 that are left. 00:07:24.120 --> 00:07:30.100 We pick one of them and put it in the third slot, and we have 00:07:30.100 --> 00:07:33.080 n minus 2 choices at this point. 00:07:33.080 --> 00:07:34.580 We continue this way. 00:07:34.580 --> 00:07:36.230 We keep going on. 00:07:36.230 --> 00:07:39.640 At some point we have placed n minus 1 of the 00:07:39.640 --> 00:07:41.090 elements into slots. 00:07:41.090 --> 00:07:44.730 There's only one element left, and that element, necessarily, 00:07:44.730 --> 00:07:47.610 will get into the last slot. 00:07:47.610 --> 00:07:50.730 There are no choices to be made at this point. 00:07:50.730 --> 00:07:57.500 So the overall number of ways that we can carry out this 00:07:57.500 --> 00:08:01.160 process, put the elements into the n slots, by the counting 00:08:01.160 --> 00:08:04.510 principle is going to be the product of the number of 00:08:04.510 --> 00:08:09.870 choices that we had at each one of the stages. 00:08:09.870 --> 00:08:13.710 So it's the product of the numbers n, n minus 1, n minus 00:08:13.710 --> 00:08:17.660 2, all the way down to 1. 00:08:17.660 --> 00:08:24.430 And this product we denote as a shorthand this way, which we 00:08:24.430 --> 00:08:26.300 read as n factorial. 00:08:26.300 --> 00:08:31.100 n factorial is the product of all integers from 1 all the 00:08:31.100 --> 00:08:32.799 way up to n. 00:08:32.799 --> 00:08:37.640 And in particular, the number of permutations of n elements 00:08:37.640 --> 00:08:40.900 is equal to n factorial. 00:08:40.900 --> 00:08:44.039 Let us now consider another example. 00:08:44.039 --> 00:08:48.460 We start again with a general set, which 00:08:48.460 --> 00:08:51.500 consists of n elements. 00:08:51.500 --> 00:08:55.120 And we're interested in constructing a 00:08:55.120 --> 00:08:56.880 subset of that set. 00:08:59.472 --> 00:09:02.350 In how many different ways can we do that? 00:09:02.350 --> 00:09:06.540 How many different subsets are there? 00:09:06.540 --> 00:09:09.700 Let us think of a sequential process through which we can 00:09:09.700 --> 00:09:11.040 choose the subset. 00:09:11.040 --> 00:09:14.580 The sequential process proceeds by considering each 00:09:14.580 --> 00:09:18.640 one of the elements of our set, one at a time. 00:09:18.640 --> 00:09:20.750 We first consider the first element, and 00:09:20.750 --> 00:09:22.650 here we have 2 choices. 00:09:22.650 --> 00:09:27.090 Do we put it inside the set or not? 00:09:27.090 --> 00:09:29.260 So 2 choices for the first element. 00:09:29.260 --> 00:09:31.600 Then we consider the second element. 00:09:31.600 --> 00:09:32.990 Again, we have 2 choices. 00:09:32.990 --> 00:09:36.910 Do we put it in the subset or not? 00:09:36.910 --> 00:09:42.770 We continue this way until we consider all the elements. 00:09:42.770 --> 00:09:45.020 There's n of them. 00:09:45.020 --> 00:09:48.020 And the overall number of choices that we have is the 00:09:48.020 --> 00:09:53.520 product of 2 times 2 times 2, n times, which is 2 00:09:53.520 --> 00:09:56.620 to the power n. 00:09:56.620 --> 00:09:59.760 At this point, we can also do a sanity check to make sure 00:09:59.760 --> 00:10:02.320 that our answer is correct. 00:10:02.320 --> 00:10:07.770 Let us consider the simple and special case where n is equal 00:10:07.770 --> 00:10:12.270 to 1, which means we're starting with this set with 1 00:10:12.270 --> 00:10:14.650 element, and we want to find the number of 00:10:14.650 --> 00:10:16.320 subsets that it has. 00:10:16.320 --> 00:10:20.820 According to the answer that we derived, this should have 2 00:10:20.820 --> 00:10:24.740 to the first, that is 2 subsets. 00:10:24.740 --> 00:10:26.830 Which ones are they? 00:10:26.830 --> 00:10:33.160 One subset of this set is the set itself and the other 00:10:33.160 --> 00:10:36.090 subset is the empty set. 00:10:36.090 --> 00:10:39.720 So we do have, indeed, 2 subsets out of that set, which 00:10:39.720 --> 00:10:42.600 agrees with the answer that we found. 00:10:42.600 --> 00:10:47.460 Notice that when we count subsets of a given set, we 00:10:47.460 --> 00:10:52.720 count both the set itself, the whole set, and we also count 00:10:52.720 --> 00:10:53.630 the empty set. 00:10:53.630 --> 00:10:56.420 All of these are subsets of our set. 00:10:58.960 --> 00:11:03.430 At this point, we can now pause and you can try to 00:11:03.430 --> 00:11:07.290 answer some simple questions of the same kind as the ones 00:11:07.290 --> 00:11:08.600 that we just practiced.