 Hi, I'm Zor. Welcome to Unizor Education. Well, you know this entire course of advanced mathematics for teenagers is actually about teaching how to solve problems. And that's why it's very important before you listen to a lecture like this, which is devoted to solving problems, you really have to try to solve them yourself. So go to Unizor.com. This lecture contains notes. Notes contain answers or hints, maybe, and try to solve these problems yourself. If you want to learn how to solve problems, you have to, guess what, solve problems, right? All right, so let's solve problems on combinatorics and they have four relatively easy problems. So let's just start. Number one. We have certain identity which I would like to prove. This is a very simple identity. It's true for any k and n where these expressions make sense, which basically means that k and n are supposed to be natural numbers and k should not exceed n. Actually, k can be zero as well. All right, so how can we prove it? Well, it's easy and it's basically a direct consequence of the formula which expresses the number of combinations from n by k or from anything else. So let's just do it. The first one is n factorial divided by k factorial and n minus k factorial. Second one is n factorial divided by k minus 1 factorial and n minus k plus 1 factorial. Well, we obviously need the common denominator. Now what's the common denominator in this case? I would like to have k. Now this is k minus 1 factorial, so I have to multiply this by k in the numerator and denominator and if k minus 1 factorial multiplied by k, it would be k factorial, right? Now here, this is 1 greater than this. So I have to multiply this n minus k factorial by n minus k plus 1 and that would become n minus k factorial. So my common denominator is this. Now this should be multiplied by n minus k plus 1 and this should be multiplied by just k. Equals n factorial times n minus k plus 1 plus k would be n plus 1, right? Equals. What is n factorial multiplied by n plus 1? That's n plus 1 factorial, right? Here I have k factorial and n plus 1 minus k, let me just change the order. n plus 1 minus k factorial. And what is this? That's exactly what it is. It's n plus 1 factorial, k factorial and n plus 1 minus k factorial. That's the end of it, very simple. All you have to know is just how to express the number of combinations in the factorial form. Number two, we have to simplify the following expression, which is 1 times p1 plus 2 times p2 plus etc plus n times pn. n is number of permutations, which is n factorial, right? Okay, how can we simplify that? Well, whenever you have this type of a series which you have to summarize, most likely, and you just have to probably remember this as a technique, each particular member can be represented as a difference between two members, a minus b, c minus b, etc. And then something would just cancel out, except the last and the first one. So that's the technique. Now, what is this type of a technique in this particular case? Well, let's just think about it. pk and k. I can always put it as k, this is k factorial, right? Now, instead of k, I will put k plus 1 minus 1 k factorial. k plus 1 times k factorial is k plus 1 factorial, right? Minus 1 times k, this is k factorial. So that's basically the representation I was looking for. That k times the number of permutations of k is equal to the difference between k plus 1 factorial and k factorial. Alright, so that's very easy then, because then what we can do is, I can write 1 times p1 would be 2 factorial minus 1 factorial, right? 1 is equal to k is equal to 1, so it's 2 factorial minus 1 factorial. Plus 3 factorial minus 2 factorial plus, etc. And the last one would be n plus 1 factorial minus n factorial. So, what's left? This is cancelling out. And what's left is n minus 1 factorial minus 1 factorial, which is 1. So that's the result. So this sum is equal to n plus 1 factorial minus 1. As you see, these are easy problems. And they immediately follow from the representation of the correspondingly combinations or permutations. And you just have to make a very, very small guess in the last problem. And actually, if you go to the notes on unizord.com for this lecture, I specify this particular approach as a hint for you. It's not a complete solution, but hint I do give. All right, next. Something similar from k to n. k times cnk equals to n times 2 to the n minus 1. Well, that seems to be a little bit more difficult. So you have to prove this formula. So k is a summation index, which basically means you start from 1. So it's 1 times c from n to 1 plus 2 times c from n to 2 plus et cetera plus n times c from n to n. All right, how can we deal with this thing? Well, basically, the idea is, well, similar, but in this case instead of converging this into some kind of a difference between two things, each one. I have to do something else, but basically I have to drive to something familiar. Now, I am not familiar with this sum, but I am familiar with this sum. Now, during the previous lecture, when I was talking about the problem of two mailmen which are supposed to deliver n different letters to n different addresses, and I was asking how can they distribute the job among them. I have suggested two different approaches. One is the following. We take the first letter and we have two choices. Give it to the first mailman to deliver or to the second one. So we have two choices. Then let's take the second letter out of n. We also have two choices, to the first or to the second, et cetera, et cetera. So we have n different letters and n different times, we have to repeat two choices. Two times two, et cetera, times two, which is two to the nth degree, right? Power of n. On the other hand, you can distribute this job differently. Well, the first mailman can take either zero letters from this n to deliver or one letter or two letters, et cetera, et cetera. So let's just count them one by one. How many different ways for this particular first mailman are there for him to take just zero letters to deliver? So all other n letters would be for the second mailman. Well, it's obviously number of combinations from n by zero. Now, how many different ways for the first man are there? So he is supposed to deliver only one letter. Well, that's how many, number of combinations from n to one, et cetera, et cetera. And the last, how many different combinations are for him to deliver all n letters so nothing would be left to the second mailman? This how many? So if you will summarize them together, you are supposed to get exactly the same number of combinations. So that's how by using two different logics to approach the same problem, I came up with this particular identity. On another hand, this is actually a manifestation of the binomial, of the Newton's binomial, which was addressed during my lectures on induction. All right, so I know this, but this is different. So the question is, can I somehow convert this into this? Well, actually, let's just think about it. What is k times c and k? Well, it's k times n factorial divided by k factorial and n minus k factorial. Right? All right, fine. But now this is k, this is k factorial. k factorial has the last member in the product of all numbers from 1 to k as a k. So we can reduce it by k. So it would be n factorial divided by k minus 1 factorial times n minus k. Now, n minus k is a difference between n and k, but it's also a difference between n minus 1 and k minus 1, right? So I can rewrite it as n times n minus 1 factorial divided by k minus 1 and n minus k factorial, which is equal to n times c from n minus 1 by k minus 1. Right? n minus 1 factorial k, this is factorial, k minus 1 factorial and the difference between them, which is n minus k. So what I have to compare this and this? So I can convert any member of this into this. Why this is better? Because these k are different and this n is a constant regardless of the number k. So what I can say is the following. I can wipe out this. I can put here this particular identity. Actually, maybe it would be pedagogically correct for me to introduce this as a separate problem. Just prove this particular identity and then introduce this problem where it can be used. Okay, in any case, what does it mean? Well, let's start from the beginning. This is k is equal to 1, so I have n c from n minus 1 to 0, right? k is equal to 1, so 1 minus 1 is 0, plus this, k is equal to 2. So it's n again, c from n minus 1 to 1, plus, etc. The last one would be n, c from n minus 1 to n minus 1. k is equal to n, right? In this particular case. So if you sum it up, it will be n times... Now look at this thing. What's the difference between this and this? The only difference is that this is n minus 1, which means this is n minus 1 here. And that's exactly what's necessary to prove. All right? That's it. All right, so the last problem... Okay, you have ten combinatorial problems to solve. Well, we solved four. Anyway, for instance, you have ten, but we solved it in one day, right? So our problems. You have to solve ten problems in three days. Now, the condition is there should be no idle days. So at least one problem should be solved in any of these three days. Well, the question is how can these ten problems be distributed among three different days in such a way that there are no idle days? No days when you solve no problems, right? Okay, that's actually quite easy. Look at this. These are ten problems. If you want to separate these ten objects into three different categories in such a way that there are no empty categories, what you have to do is you have to use these gaps. So you have ten objects. So you have nine gaps. You have to pick basically any two gaps. Let's say this one and this one. Now, these two gaps are separating my ten objects into three different groups. And since you are choosing only these gaps in between the objects, you are assured that there will be no categories with no objects among them, because you are using only these gaps. So you have nine gaps and you have to choose two of these. So the answer is number of combinations from nine to two, which is nine times eight divided by one times two, which is thirty-six. All right, that's the last problem. And as usually, I recommend you to go to the notes for this lecture on Unisor.com. And you have these problems in the notes. Try to solve them just yourself completely. Check the answers and if you are satisfied, that's great. It's very important to start and if you cannot solve it in the beginning to finish this particular work with solving the problems yourself. That's very important. Now, in addition, as I'm sure most of you are aware, if you are a registered student, which you can register just with your ID, basically, and password, then you can take exams. Exams are also very important. Not every topic has exams. For instance, for combinatorics, I didn't really put yet exams, but I will. And to do that, you have to really have registered yourself. And there should be another person who is like a supervisor or a parent who also registers as a supervisor or a parent. And that person is supposed to enroll you into a certain topic or maybe on the top you can be enrolled in the whole course of advanced mathematics. Then whatever you are enrolled in will be presented to you in the same basically order as you are having this information on the open side of the Unisor.com. But it actually will be available with exams which you can take and it's a very good way to basically check yourself. Well, that's it for today. Thank you very much and good luck.