 Hi, I'm Zor, welcome to Unizor Education. I'd like to start a brand new topic, combinatorics. And the first part is related to a concept called permutations. Well, generally speaking about combinatorics, I like this particular topic. It's quite interesting in one aspect. If you're talking about, let's say, algebra solving equations, well, either you solve it or you don't. Geometry, proving a theorem. I mean, either you go logically step by step and you basically prove it or you don't. In combinatorics, the problem is calculate something. And you're thinking that you're approaching the problem correctly and you get some answer and it does not correspond to the answer in the textbook, let's say. So you're thinking differently and, well, you get the different result, which also might or might not correspond to the textbook answer. So my point is that in combinatorics, your mind should be very, very sharp to, like, think about all the possibilities, all the different variations or whatever. So all the incorrect answers to the problems are lying quite closely to the correct ones. So if in algebra either you solve it or you don't, they are very much far apart. In combinatorics, correct and incorrect answers might be very close. And every time you think about the problem slightly differently, you might get the different result. So that's why it's very important to go through lots of problems, make a lot of mistakes, quite frankly, and gradually develop your sense of how to solve the combinatorics problems correctly. It takes time and practice. Well, I'm planning to do more different problem solving in this particular chapter of the partition of this course than usually, just because this is the most important part of this part, the problem solving. Because the definitions, if I will give you the formula in definitions, etc., that's simple. To solve many, many problems which really use this type of logic, these formulas or whatever to come up with some correct result, that takes practice. And as I was saying, it's not very easy to differentiate between the right and wrong. So that's why your mind should be very much sharpened to do this type of thing. So today is permutation. So first of all, generally speaking, what permutation is. Permutation is a task of putting a certain number of objects in certain order. For instance, you have to go to three different places, let's call it A, B, and C. You have to visit them. Let's say it's a mother's day and you have a grandmother, another grandmother, and the wife to basically congratulate with the mother's day or something like this. So question is, which order you prefer to do this? Well, you can do it in order A, B, C, or you can do it in order A, C, B. There are no more different orders when A is the first one. So after A it's either B, C, or C, B. Now, on the first place, you can visit B. And then you can have, again, two different things, A, C, or C, A. Finally, your first choice may be the place C, after which you have two different variations. These are all different orders. Three objects, A, B, and C, can be put in sequence one after another. All right, so it's six, right? Six different variations. Now, let's get a little bit more general problem. Let's say you have N different objects. So you have N different objects. Well, let's think about it this way. You have, let's say, a pile of N objects, and you would like to put them in line. For instance, let me just give you another example before this. If you are in command of a group of soldiers and you have to put them in line, you can put them in line by their height, let's say, order them by height, from tallest to the shortest, or from shortest to the tallest, or alphabetically by their last names, or by color of their hair, from dark to light, or from light to dark, I mean, in any order. So these problems really occur. The question is how many different ways of putting this number of soldiers into a line can be. So let's get to the general problem of N objects, which are just piling somewhere, and you are putting them in order. Well, for the element number one in our ordered set of N objects, we have a choice to make. We have N different objects. So for number one, we have N different choices, right? Okay, so let's just write it down. So N different choices. Now, after we have made a choice for a position number one in our row, we have to make a choice for the second. Now, there are N minus one elements left in this pile, right? Each one of them can be number two. So it looks like for every choice we made for number one, and there are N of these choices, we have N minus one choices for object number two. And for each pair of two first objects, we have how many are left? N minus two, right? We have N minus two choices for the position number three. So we are continuing this process, and we are multiplying the number of our choices, because with every number of choices we made already, we still have so many choices to make for the next step. So it goes up to the very last one. So the one before last we had two choices left, and then there is only one. So it looks like this might be the formula for the number of permutations of N objects. The number of ways we can order these N objects into one row. By the way, permutation usually is symbolically designated as this. All right, now, this is where many textbooks are actually ending the explanation about what is the number of permutations, they give this formula. And by the way, this formula has a shorter notation, N factorial. N factorial is the product of all natural numbers from one to N. N is assumed to be a natural number, integer positive. So for every integer positive number, N factorial is the product of all the numbers from one to this number, which is exactly this one. Obviously, it doesn't really matter the order from N to one or from one to N because the multiplication is commutative. So in most cases, this is the formula where textbooks are ending their explanation. I would like actually to move a little bit further. We are trying to make this course as rigorous as possible. In this particular case, I'm just explaining basically in words how the permutation goes and the number of different choices to make, etc. It's not a really mathematical proof and I would like to prove it more or less rigorously, mathematically, so to speak. All right, so how can I prove that the number of permutation is such and such? Well, the perfect way to prove this is induction, mathematical induction. So let's just think about it. What if I have, now to prove by induction, I have to do three steps, right? Check if the formula is correct for some initial number, like N equals to one, for instance. Then assume that for certain number N equals to k formula is correct and then derive from this that the formula is correct for the next N is equal to k plus one. So that's what I'm going to do. All right, step number one is to check that the permutation of one element is exactly one factorial. Well, one factorial is one and a product of all numbers from one to one, which is one. And permutation of one element. Well, how many times, if I have only one element, how many times I can put it in order? Well, there is no other element, so there is only one way. This is element number one and the first and the last element, right? So this formula actually is correct for N is equal to one. All right, second. Let's assume that for some N equal to k, the formula is correct. Now, let's check what happens for the next... So let's assume we have k plus one objects. What I will do, I will separate this pile of k plus one objects into k, first k, if you wish, and separately one. So I'll just pick up one object, doesn't matter which one, just pick it up. So now I have a group of k objects in my pile. And to make the whole thing in order, I will first order in some way my remaining k objects. And then I will put this k plus first object somewhere among the first ks, right? If I have k objects which are remained in the pile after I extracted one, this, by our assumption, I can put these k different objects into an order in k factorial different ways, right? That's our assumption. Let's assume that this is some way we have placed our k objects. In this case, k is equal to four. And now I have one more object which I picked up from the pile before. I can put it here, I can put it here, here, here, and here. Each way doesn't really matter where exactly I put it. But every time I put it, I will get a certain row of k plus one objects, right? I can put it here, here, here, and here. And this is my k objects, right? And with this one, it will be k plus one. So, as soon as I did that from the combination which I had, I will obtain a new combination of all k plus one objects, right? How many? Well, with each combination of the k objects, I have additionally how many of those guys, how many places? k plus one, before first element, from first to second, from second to third, etc. From k minus one to k, and then after k, right? So, if these are k objects, there are k plus one places where I can put my k plus first object. Before the first, in between, and after the last. So, it looks like with every combination of the k objects, I have k plus one different combinations of k plus one objects, right? Because for each one of these, I can have this k plus one objects placed in k plus one different places. So, with each of these, I have k plus one new positions of the k plus one objects. So, it looks like my formula for number of permutations of k plus one elements should be equal to number of permutations of the k elements. And since for each of these guys, I have k plus one new permutations of the total. So, that is supposed to be the true statement. And obviously, if k factorial is a pk, k factorial times k plus one, it's actually k plus one factorial, right? Because k plus one factorial is the product of all natural numbers from one to k plus one, which is from one to k and k plus first, right? So, that actually proves that the formula is exactly the same for k plus one. So, this is a proof by induction, which I think should make you feel a little better, because we have rigorously, well, as rigorously as we could actually, prove this validity of this formula and instead of just giving it to you based on certain philosophical discussions, right? But anyway, the formula is very simple. This factorial thing is really very helpful to shorthand the product of all the numbers from one to this particular given number. Anyway, this is the final result I wanted to present to you. The permutations of certain number of objects, number of permutations. Now, I assume that objects are different because if the objects are the same, you will not be able to differentiate one permutation from another. And this is yet another problem in combinatorics, which we will address. But right now, we are talking about different objects and different objects which are positioned in certain order and the question is how many different orders of positioning of these and objects exist, and the answer is n factorial. Okay, I do recommend you as usually to use the Unizor website, not just the YouTube lecture. And the website contains notes for this and all other lectures. And also it provides you the ability to sign in and take exams. All it requires is actually to have some kind of a supervision and enrollment into the courses. Everything is completely free. So you're welcome. I do recommend to read again the notes for this lecture. And it has this proof by induction. I think it's always helpful to refresh your memory. So thanks very much and good luck.