1 00:00:00,260 --> 00:00:01,540 Welcome back. 2 00:00:01,540 --> 00:00:06,790 So we have been looking at how to count the cardinality of sets. 3 00:00:06,790 --> 00:00:13,140 So this is what we call as Combinatorics, so which is the branch of maths that involves 4 00:00:13,140 --> 00:00:15,440 counting. 5 00:00:15,440 --> 00:00:21,940 So most of the time the typical question is, given a set S what is the cardinality of S, 6 00:00:21,940 --> 00:00:28,810 or in other words what is the size of the set S, or how many elements are there in the 7 00:00:28,810 --> 00:00:31,170 set S. 8 00:00:31,170 --> 00:00:38,780 Now the question to be asked is that how is the set given, and most of the time when we 9 00:00:38,780 --> 00:00:45,999 talk about this kind of problem, we say that the set is given implicitly. 10 00:00:45,999 --> 00:00:50,850 It is not given explicitly, it has been described. 11 00:00:50,850 --> 00:00:57,940 For example, it can be something like, how many elements of a particular universe set 12 00:00:57,940 --> 00:01:00,690 satisfy certain conditions. 13 00:01:00,690 --> 00:01:08,510 Or Equivalently put, how many ways can you draw an element from the set, such that the 14 00:01:08,510 --> 00:01:10,420 element satisfies a set of conditions. 15 00:01:10,420 --> 00:01:19,400 Now we have been, we have some problem that we should keep in our mind before we what 16 00:01:19,400 --> 00:01:21,610 we should like to solve. 17 00:01:21,610 --> 00:01:28,080 The first one being, how many n digit numbers are there, in decimal representation with 18 00:01:28,080 --> 00:01:34,390 no consecutive digits same. 19 00:01:34,390 --> 00:01:44,920 The second, how many functions are there from the set one to n to one to k, that are not 20 00:01:44,920 --> 00:01:45,920 decreasing. 21 00:01:45,920 --> 00:01:52,550 The third, how many ways to distribute n identical toffees among k kids. 22 00:01:52,550 --> 00:01:59,659 And the last one is the number of 01 string of length n, which does not have any consecutive 23 00:01:59,659 --> 00:02:00,890 zero. 24 00:02:00,890 --> 00:02:13,680 Now, question is that, how to count is something an extremely challenging problem. 25 00:02:13,680 --> 00:02:22,459 Every problem, every set that we have to count has its own complications and should be tackled 26 00:02:22,459 --> 00:02:23,459 differently. 27 00:02:23,459 --> 00:02:32,569 Even, some of the greats of math like, Srinivasa Ramanujan has worked on counting for quite 28 00:02:32,569 --> 00:02:35,769 a big part of his life. 29 00:02:35,769 --> 00:02:42,370 And, here in this few lectures we will be giving some tricks and tools, which will help 30 00:02:42,370 --> 00:02:44,409 you to attack problems. 31 00:02:44,409 --> 00:02:53,730 But again, as I told you, as I tell in every class, you should use your creative mind to 32 00:02:53,730 --> 00:03:00,260 understand, which problem can be solved using which trick, every problem is different, every 33 00:03:00,260 --> 00:03:02,230 problem is unique. 34 00:03:02,230 --> 00:03:11,419 Now, Till now, we have been looking at, we have already looked at this problem of how 35 00:03:11,419 --> 00:03:20,239 many ways can I select k objects from n objects, and we have looked at the two cases, namely 36 00:03:20,239 --> 00:03:26,750 what happens whether, are we allowed to pick a same object multiple times, is repetition 37 00:03:26,750 --> 00:03:32,890 allowed, among the k selected objects. 38 00:03:32,890 --> 00:03:39,139 And, is the ordering of the objects in this k objects matter. 39 00:03:39,139 --> 00:03:44,959 And we have seen how to solve all the four cases. 40 00:03:44,959 --> 00:03:54,980 The next one that we looked at was, how many ways can we distribute n balls into k bins. 41 00:03:54,980 --> 00:04:04,569 And there are few cases to be studied, particularly whether the bins are distinguishable or not, 42 00:04:04,569 --> 00:04:09,519 whether the balls are distinguishable or not. 43 00:04:09,519 --> 00:04:19,630 If the balls are distinguishable, is the ordering in the bins matter, can the bins be empty, 44 00:04:19,630 --> 00:04:20,810 and is there any other restrictions that can be asked. 45 00:04:20,810 --> 00:04:27,940 And, in the last video, we looked at this problem and we solved some of the parts of 46 00:04:27,940 --> 00:04:29,230 this problem. 47 00:04:29,230 --> 00:04:44,670 Namely we did solve the case when the bins are labelled, and items are indistinguishable, 48 00:04:44,670 --> 00:04:48,420 and in case bins are labelled, but bins can be empty. 49 00:04:48,420 --> 00:04:57,190 We also solved, when the items are distinguishable and ordering matters or does not matter, either 50 00:04:57,190 --> 00:04:58,310 case we solved that. 51 00:04:58,310 --> 00:05:04,280 But we are left with two of the problems, which we will be tackling in this video, namely 52 00:05:04,280 --> 00:05:12,930 if bins are labelled but cannot be empty, then how do you solve it. 53 00:05:12,930 --> 00:05:19,310 And also we were left with the case when bins are unlabelled. 54 00:05:19,310 --> 00:05:29,090 Now, lets to before start working on those things, let us do a little bit study of the 55 00:05:29,090 --> 00:05:31,470 theory behind it. 56 00:05:31,470 --> 00:05:39,000 The important thing is that, if A and B are two sets and size of A is p and size of B 57 00:05:39,000 --> 00:05:47,050 is q, what is the size of A union B. Note that, this is very similar to the additive 58 00:05:47,050 --> 00:05:50,300 law that we studied two videos ago. 59 00:05:50,300 --> 00:05:59,230 It told that if A and B does not have anything in common or in other words, what in told 60 00:05:59,230 --> 00:06:10,300 was that if A intersection B is empty,meaning if the cardinality of this one is zero, then 61 00:06:10,300 --> 00:06:19,830 the cardinality of A union B, either A happens or B happens is p plus q. 62 00:06:19,830 --> 00:06:29,940 But what happens if this is not zero, for example, if I have this as the set, so if 63 00:06:29,940 --> 00:06:42,060 this is universal set Z, where we are working on, if this is A and this is B, and A intersection 64 00:06:42,060 --> 00:06:47,980 B does have some place here, how do you solve it. 65 00:06:47,980 --> 00:06:57,030 Now as you can see, from the picture in some sense, what happens if I look at size of A 66 00:06:57,030 --> 00:07:06,550 plus size of B. So in the case that, okay, let me just complete this line, size of, B 67 00:07:06,550 --> 00:07:15,390 is this set, and A is this set. 68 00:07:15,390 --> 00:07:25,520 Now note that every point here inside A union B has been selected once, if it is somewhere 69 00:07:25,520 --> 00:07:30,710 here, and once if it is somewhere here. 70 00:07:30,710 --> 00:07:37,349 But if it in the intersection, it has been counted twice, right. 71 00:07:37,349 --> 00:07:45,310 So it is like, if you have to place two round paper cuts, one over the other, you need to 72 00:07:45,310 --> 00:07:52,030 cut away, one should cut away this chunk out once. 73 00:07:52,030 --> 00:07:59,900 So, if you subtract A intersection B, you should be getting A union B. 74 00:07:59,900 --> 00:08:07,420 And that is a very crucial thing, that A union B is size of A plus size of B minus intersection 75 00:08:07,420 --> 00:08:23,070 of A and B. Now, so what is the answer, answer is of course that A union B is size of A plus 76 00:08:23,070 --> 00:08:25,820 sign of B minus A intersection B. 77 00:08:25,820 --> 00:08:35,550 Now, what happens if I have three sets, we are back to the same Venn diagram for set 78 00:08:35,550 --> 00:08:47,960 we always look at for such kind of situations, if we have A, B and C, then you can first 79 00:08:47,960 --> 00:08:54,260 pick up all of them, you can first cover all of them. 80 00:08:54,260 --> 00:09:17,240 But then, say this is A, this is B and this is C. Now As you can see, that this space 81 00:09:17,240 --> 00:09:27,210 of A intersection C has been counted twice, first by A and then by C. So i need to subtract 82 00:09:27,210 --> 00:09:30,820 at least once one of this away, right. 83 00:09:30,820 --> 00:09:38,930 So subtract A intersection C, and also I have to similarly subtract A intersection B and 84 00:09:38,930 --> 00:09:51,480 I have to subtract A intersection B, sorry I have to subtract B intersection C. Now, 85 00:09:51,480 --> 00:10:03,480 what happens in this case of this small area, this was counted once for A, once by B, once 86 00:10:03,480 --> 00:10:12,230 by C. It was subtracted once by A intersection C, subtracted once by A intersection B, subtracted 87 00:10:12,230 --> 00:10:18,200 once by B intersection C, that means, right now this one has not contributed anything. 88 00:10:18,200 --> 00:10:26,500 So I need to now add it up, so I have to add this part, A intersection B intersection C. 89 00:10:26,500 --> 00:10:39,350 So in other words what I will get is that, I will get that A union B union C is equal 90 00:10:39,350 --> 00:10:47,510 to first adding the sets individually, then subtracting the pairwise intersection then 91 00:10:47,510 --> 00:10:48,970 adding the threewise intersection. 92 00:10:48,970 --> 00:10:59,700 Now, question is that what happens in the case of 4 case, if I have four sets A, B, 93 00:10:59,700 --> 00:11:07,250 C, D, in that case what is the size A union B union C union D. 94 00:11:07,250 --> 00:11:18,160 Is there a formula for it and we do have one and that is called the principle of inclusion/exclusion 95 00:11:18,160 --> 00:11:25,120 and it basically talks about that if I have n sets A1 to An, then what is the size of 96 00:11:25,120 --> 00:11:29,250 A1 union A2 union till An. 97 00:11:29,250 --> 00:11:42,660 So the size of the union is first add the individual sets, then subtract all the pairs 98 00:11:42,660 --> 00:11:48,900 of intersections, then add all the triple intersections, then subtract all the four 99 00:11:48,900 --> 00:11:50,310 intersections. 100 00:11:50,310 --> 00:11:59,960 In some sense you are adding, subtracting, adding, subtracting and it keeps on going 101 00:11:59,960 --> 00:12:08,100 on till the last one where you add or subtract depending on what is the size of n, if n is 102 00:12:08,100 --> 00:12:13,210 odd or even, it is minus one power n plus 1. 103 00:12:13,210 --> 00:12:18,922 You add or subtract the intersection of all the sets. 104 00:12:18,922 --> 00:12:20,840 Now this is a theorem. 105 00:12:20,840 --> 00:12:25,990 I am not going to prove this theorem. 106 00:12:25,990 --> 00:12:31,220 I leave this to you guys to check that this theorem is indeed correct. 107 00:12:31,220 --> 00:12:34,800 You can prove this theorem using induction. 108 00:12:34,800 --> 00:12:41,710 This particular theorem is an extremely powerful tool for counting. 109 00:12:41,710 --> 00:12:49,690 So writing is in a nice way, we do get this particular expression. 110 00:12:49,690 --> 00:13:01,040 So once we have this principle of inclusion/exclusion, we can use it to solve a many of our problems. 111 00:13:01,040 --> 00:13:04,430 So let us start with this problem. 112 00:13:04,430 --> 00:13:13,120 Where We say that there are n constables in a police station. 113 00:13:13,120 --> 00:13:19,230 You have to divide it into four projects, project 1, project 2, project 3, project 4. 114 00:13:19,230 --> 00:13:24,570 No constable can be part of more than one project and every constable should be part 115 00:13:24,570 --> 00:13:28,810 of some project and also at least one constable should be assigned to every project. 116 00:13:28,810 --> 00:13:33,070 None of the project should be empty. 117 00:13:33,070 --> 00:13:37,730 Now, how do you solve it? 118 00:13:37,730 --> 00:13:45,700 If you remember, if we do not have this particular thing that every project has at least one 119 00:13:45,700 --> 00:13:52,700 constable assigned, then we saw that the first constable can be put in either of the four 120 00:13:52,700 --> 00:13:59,220 projects, the second constable can be put in either of the four projects, and so on, 121 00:13:59,220 --> 00:14:07,190 so the total number of ways of assigning the constable in the four projects is four power 122 00:14:07,190 --> 00:14:09,649 n, right. 123 00:14:09,649 --> 00:14:21,200 But here unfortunately, it is a possibility that some of the projects might be empty. 124 00:14:21,200 --> 00:14:23,730 So we need to get rid of them. 125 00:14:23,730 --> 00:14:33,000 If we want to ensure that every project have some constable, then we have to subtract all 126 00:14:33,000 --> 00:14:45,500 the possible, so subtract all distributions where at least one project is empty. 127 00:14:45,500 --> 00:15:02,950 Now, how do you do it. 128 00:15:02,950 --> 00:15:05,529 So let me define this new class Ai. 129 00:15:05,529 --> 00:15:09,900 Ai is the set of all distributions. 130 00:15:09,900 --> 00:15:18,910 Distribution meaning of n constables into projects where project i is empty or it does 131 00:15:18,910 --> 00:15:23,670 not have any constable. 132 00:15:23,670 --> 00:15:42,899 Now if I have defined this Ai in this way, then how do I define the set of all distributions 133 00:15:42,899 --> 00:15:44,029 that are empty. 134 00:15:44,029 --> 00:15:58,899 So the set of all distributions that are empty is nothing but union over Ai or rather in 135 00:15:58,899 --> 00:16:08,220 this case it will be A1, all the distribution where the first project is empty or all the 136 00:16:08,220 --> 00:16:12,560 distribution where the second project is empty, or all the distribution where the third project 137 00:16:12,560 --> 00:16:19,700 is empty, or all the distribution where the fourth project is empty and substract them 138 00:16:19,700 --> 00:16:21,180 out, right. 139 00:16:21,180 --> 00:16:29,230 So we can now subtract, so this is the set of all assignments, or all distributions of 140 00:16:29,230 --> 00:16:38,570 constables in to four projects where some of or at least one of the project is empty 141 00:16:38,570 --> 00:16:46,360 and when I subtract that one from four power n, I get my required answer. 142 00:16:46,360 --> 00:16:56,110 Now by the principle of inclusion/exclusion, I can write this one as, so all I have to 143 00:16:56,110 --> 00:17:01,450 do is that I have to write this particular term, right, the unions. 144 00:17:01,450 --> 00:17:03,959 So let me just do not worry about this one. 145 00:17:03,959 --> 00:17:06,740 This one, what is this term? 146 00:17:06,740 --> 00:17:14,199 This is summation of A1 plus A2 plus A3 plus A4. 147 00:17:14,199 --> 00:17:15,630 Now what is A1 plus A2 plus A3 plus A4? 148 00:17:15,630 --> 00:17:18,350 What is A1? 149 00:17:18,350 --> 00:17:25,260 A1 is the number of ways in which the i-th project is empty. 150 00:17:25,260 --> 00:17:27,839 That means nobody goes to the i-th project. 151 00:17:27,839 --> 00:17:33,840 How many such ways can you distribute n constable into four projects, where the first project 152 00:17:33,840 --> 00:17:34,840 is empty. 153 00:17:34,840 --> 00:17:44,840 That means it is the same way as distributing n constable into the project 2, 3, or 4. 154 00:17:44,840 --> 00:17:47,299 So this one is equals to 3 power n. 155 00:17:47,299 --> 00:17:56,241 The first one, The constables can go to project 2, 3 or 4, Similarly, this is also 3 power 156 00:17:56,241 --> 00:17:57,241 n. 157 00:17:57,241 --> 00:18:08,290 Now, by principle of inclusion/exclusion, I have to subtract A1 intersection A2, I have 158 00:18:08,290 --> 00:18:11,650 to subtract A2 intersection A3 and so on. 159 00:18:11,650 --> 00:18:15,360 Now what is A1 intersection A2? 160 00:18:15,360 --> 00:18:26,360 A1 intersection A2 is the set of all distributions where the first and second project are empty, 161 00:18:26,360 --> 00:18:31,320 right and this can of course happen with that both the projects are empty, that means it 162 00:18:31,320 --> 00:18:38,750 is same as dividing n constable into only project 3 and project 4, which is 2 power 163 00:18:38,750 --> 00:18:39,750 n. 164 00:18:39,750 --> 00:18:48,890 And all of them are same and similarly when we have to do it for A1 intersection A2 intersection 165 00:18:48,890 --> 00:18:54,720 A3, these are the number of distributions where project 1, project 2, and project 3 166 00:18:54,720 --> 00:19:05,580 are all empty and this can only happen with one way where all of them goes to project 167 00:19:05,580 --> 00:19:08,040 4 and so on. 168 00:19:08,040 --> 00:19:18,620 Note that this is what, so the main idea is that intersection of Ai can be sometimes very 169 00:19:18,620 --> 00:19:21,560 easy to compute as in this case. 170 00:19:21,560 --> 00:19:26,750 And if intersection of Ai is easy to compute, then by the principle of inclusion/exclusion, 171 00:19:26,750 --> 00:19:36,460 I can write cardinality of the union of the sets and by doing so, we can get the number. 172 00:19:36,460 --> 00:19:44,890 It will not be a very compact number, but it is still something, which has a expression. 173 00:19:44,890 --> 00:19:52,560 So infact we can write down the expression of that exactly. 174 00:19:52,560 --> 00:19:59,580 So if there are n objects n distinguishable ball and k labelled bins, how many ways can 175 00:19:59,580 --> 00:20:08,670 we place n balls in k bins when ordering inside the bin does not matter, but no bins can be 176 00:20:08,670 --> 00:20:13,460 left empty, it is k power n minus this way. 177 00:20:13,460 --> 00:20:21,490 Now why this k choose i because k choose i is the number of ways in which I can choose 178 00:20:21,490 --> 00:20:25,549 i of the sets and take intersection of them, right. 179 00:20:25,549 --> 00:20:36,890 I have to for example this should be like k power n minus intersection of all the A1, 180 00:20:36,890 --> 00:20:38,120 A2 is right. 181 00:20:38,120 --> 00:20:51,600 So all the A1, A2 is k choose 2 times k minus 1 power n plus all the triples k choose 3 182 00:20:51,600 --> 00:20:59,600 and all the triples with value Ai intersection Aj intersection Ak is k minus 2 power n and 183 00:20:59,600 --> 00:21:01,930 so on. 184 00:21:01,930 --> 00:21:09,520 So this gives us a kind of a close form expression, not exactly a close form expression, but an 185 00:21:09,520 --> 00:21:20,169 expression of this kind for getting the number of ways in distributing n balls into k bins 186 00:21:20,169 --> 00:21:23,080 with no empty bin. 187 00:21:23,080 --> 00:21:33,460 So we can use the same technique, when ranking inside the bins matter. 188 00:21:33,460 --> 00:21:42,429 If you remember, ranking inside the bin without when they are empty bins were allowed, was 189 00:21:42,429 --> 00:21:43,429 this one. 190 00:21:43,429 --> 00:21:51,679 Now, we can do the identically same thing with the principle of inclusion exclusion 191 00:21:51,679 --> 00:21:57,220 and we can get this. 192 00:21:57,220 --> 00:22:06,360 So by doing so, we have been able to fill up this particular matrix. 193 00:22:06,360 --> 00:22:16,770 Now let us try to see if we can solve the other three possibilities, namely when bins 194 00:22:16,770 --> 00:22:17,770 are unlabelled. 195 00:22:17,770 --> 00:22:19,460 Now, when bins are unlabelled for example. 196 00:22:19,460 --> 00:22:32,590 This is a typical example when you have to the divide constable into four groups and 197 00:22:32,590 --> 00:22:42,919 we do not care about which group is assigned to which project, how many ways can you do 198 00:22:42,919 --> 00:22:43,919 it? 199 00:22:43,919 --> 00:22:53,630 Now the idea is that if say I assign constable C1, C2, C3 to project 1 and C4 to project 200 00:22:53,630 --> 00:23:07,299 2 and C5 to project 3 and C6, C7 to project 4, note that if I permute this project 1 to 201 00:23:07,299 --> 00:23:13,200 project, if you do not have any name for the projects. 202 00:23:13,200 --> 00:23:27,490 In that case if I change the project ordering to P2, P3, P1, P4, any of the ordering are 203 00:23:27,490 --> 00:23:34,510 equivalent and they would have been counted in when we did not take care of the fact that 204 00:23:34,510 --> 00:23:36,710 they are not indistinguishable. 205 00:23:36,710 --> 00:23:49,720 By using this technique of the equivalent classes, we can see that every distribution 206 00:23:49,720 --> 00:24:00,480 with the labelled projects has belongs to an equivalent class of the same size namely 207 00:24:00,480 --> 00:24:04,940 the number of ways you can permute these four things, which is four factorial. 208 00:24:04,940 --> 00:24:13,640 So by the technique of this equivalence partitioning, we can now just divide. 209 00:24:13,640 --> 00:24:26,172 So if I have to solve this thing of n distinguishable balls into k unlabelled bins, we have to take 210 00:24:26,172 --> 00:24:36,280 in the case when I have exactly k bins that are filled, no empty, and I divide it by k 211 00:24:36,280 --> 00:24:38,480 factorial and I get the answer. 212 00:24:38,480 --> 00:24:48,870 Note that, if I have not taken the number here, when everything was not empty, then 213 00:24:48,870 --> 00:25:00,150 things could have been a bit more weird. 214 00:25:00,150 --> 00:25:07,320 So by the same argument, by the way, this particular number, so this is the number of 215 00:25:07,320 --> 00:25:18,100 ways in which one can distribute n distinguishable balls into k unlabelled bins when the ordering 216 00:25:18,100 --> 00:25:28,230 in the bins does not matter, but no bins can be left empty. 217 00:25:28,230 --> 00:25:40,260 This is called the Sterling Number of Second Kind denoted by S (n, k). 218 00:25:40,260 --> 00:25:53,870 Similarly, we can do it for the third, for the last of this table, which is when ordering 219 00:25:53,870 --> 00:26:01,780 inside matters does not matter and bins are unlabelled. 220 00:26:01,780 --> 00:26:12,850 Now with this, we have done basically all the cases except this case namely when items 221 00:26:12,850 --> 00:26:18,750 are indistinguishable and bins are unlabelled. 222 00:26:18,750 --> 00:26:21,630 How many ways can you do it? 223 00:26:21,630 --> 00:26:30,190 Now, This is a very hard problem and we will be talking about this problem in the next 224 00:26:30,190 --> 00:26:31,190 video. 225 00:26:31,190 --> 00:26:32,170 Thank you.