 Hi, I'm Zor, welcome to Unizor Education. This lecture is part of the whole course of advanced mathematics presented on Unizor.com and that's where I suggest you to watch this lecture because it has comments over there, the notes basically, and it's very important this particular lecture and many others dedicated to solving problems and I very strongly recommend you to try to solve all these problems just by yourself. So on the Unizor.com website, among the notes for this lecture, all these problems are presented with answers and solutions, written solutions. So you solve the problem yourself, try, check against the answer and then you can read the solution and then I suggest you to listen to this lecture where I basically explain it in more details. Alright, so this is a second series of problems dedicated to a little bit more advanced levels of combinatoric. Well, just a little, I mean it's really not very difficult problems at all. Alright, so let me just go straight ahead with four different problems and the problem number one is you have m objects and n objects. They're all different, all different objects. Now, the question is how many different groups of objects which contain x objects from the m sets and y objects from the group of n objects positioned in certain orders, the order is important. So again, we take x objects from the first group of m objects and y objects from the second group of n objects. We position them in some order, all these x plus y objects. So the question is how many different sequences of x plus y objects you can get. Alright, so basically the problem can be reformulated in the following way. Well, how many different groups of x objects from these m can be obtained? Well, we write it down and how many different groups of y objects of these n, we write it down. Now, how many times we can basically permutate them to get certain different sequences of these objects? So that's basically the formulation and the plan to solve the problem. So the first number is how many different subsets of x elements out of m exist? Obviously, number of combinations from m by x. Now, with each of them, there are certain number of combinations of y objects out of m. Now, each of that can be combined with each of those, so that's supposed to be multiplied. Now, when we have a set of x objects from here and y objects from here, we really can put it in some sequence, right? So we have x plus y objects and we can actually have x plus y factorial different sequences, right? So that's the answer to the problem. So that's number of subsets of x elements out of m, number of subsets of y elements out of n, and then we permute all these x plus y objects and we are multiplying them because every one of them is valid with everyone else. So that's the end of this program. Next, you have n points, in this case n is equal to 5 on the plane. Now, some of them, k points, in this case, k is equal to 3, are lying on the same line, on the same straight line. Now, consider that no other line which connects these points, no lines are parallel to each other and no three lines are intersecting in the same point. So every pair of lines give you one and only one intersection point, okay? Now, question is how many lines exist that connect the pairs of these points? Okay, now let's think about it. What is the line which connects two points? Basically, you take out of these n points two in some way and you draw the line. So how many lines? As many as many pairs in theory, right? So we can start with this number, number of combinations of two points out of n. Now, however, since these k points are lying on the same line, all the different combinations which include pairs from these k are actually the same. So what I will do is I will subtract the number of all the different combinations of two points out of these k and just add one line which is this one. That actually gives me the total number of different lines which can be drawn to connect all the points, all the pairs of points, right? So it's all the different pairs minus all the pairs which are on the same line and then plus one which is this line itself. You can consider actually a different way to do this. You can just divide all the lines into three categories. Lines which connect only these points, now there are n minus k of these points, right? And the number of lines which connect them is number of combinations from n minus k by two. That's number of pairs, right? Now then there are certain number of lines which connect all these points with all these points. So any of these and there are n minus k of them can be connected with any of these and there are k of them. So that's my number. So first this is number of lines which connect only these points, then lines which connect this is this, this one, number of combinations from n minus k by two, then number of lines which connect a point from here and the point of there and this is n minus k times k because n minus k of these and k of these and then one line extra which is this. So I've got two different formulas. That's actually very good because you don't have a good way to check the results of the solution to your combination problem unless you can approach it with two different ways. This is a perfect example when you can really do this. Well, let me see if this is exactly the same. Well, n combinations of n by two is n times n minus one by two, right? Right. Minus k k minus one by two plus one. That's this number. Well, let me just have the common denominator. It would be n square minus n minus k square plus k plus two divided by two, right? n square minus minus n, k square minus k, but there is a minus in front of it and one becomes two. I think that's what it is. All right. How about this one? This one is n minus k n minus k minus one over two. That's this one. Plus n minus k by k plus one, all right? Which is equal to n square minus k n minus k n plus k square minus n plus k. That's this one. Now, this, but there is a two here, so I have to multiply by two. n k, so it's plus two n k minus two k square and plus two. All right, let's see. k square and minus two k square, that's minus k square, which is fine. k n, k n and plus two k n, these are cancel out. n square n square minus n minus n plus k plus k. k square minus two k square is minus two k square and plus two plus two. Exactly the same thing. So that's the checking and this is the perfect way to check the answer to a combination problem, the combinatorics problem, all right? If you can really approach it from completely different ways and formulas are completely different as you saw, but you have the same result after you simplify the formulas. So I'm absolutely sure this is the correct decision. All right, now next one. You have three pens, one book, one calculator and one notepad. So how many? Six objects altogether. What you're looking for is the number of ordered groups of four objects out of these six. So you pick four objects out of these, put in some sequence and that's what you need to know. Now, three pens are identical and that's the complication. You see, if these were just six different objects, that would be just partial permutation of four out of six. But this is not a typical problem so we have to somehow approach it differently. All right, but what I can suggest is the following. In our group of four objects must be at least one pen, right? Because there are only three objects of other types. So we need four, so we need at least one pen. So we have three different cases. We have one pen in a group, second case is two pens in a group and third case is three pens in a group. And let's calculate separately which one, how many different ordered sets of four objects are there with either one or two or three pens. Okay, with one pen, and I know that I have four objects, right? I have to include all these three. So I will have four different objects, one pen, one book, one calculator and one notepad. How many different ordered sets can be obtained from four different objects? Well, four factorial, right? So that gives me four factorial number of combinations. Okay, now let's do an easier way, the three. So I have three, then I have either this one or this one or this one, right? So I have three different cases. And after I choose whether the fourth one is a book or a calculator or a notepad, and I have four places, I can put any one of these into any of these four places. So it looks like it's 12, right? So any one of these can be on any of the four places. And the rest, the other three places are taken by the three pens which are identical, right? So that's what characterizes the combination. If you have three pens and one of these, then one of these can be on one of the four places. That's why the combination of these three by four places gives me 12. Now with two pens, it's a little bit more difficult. So I know that there are two pens, right? Now, first what I have to do, I have to choose which two out of these three can be presented in my set. Well, obviously it's number of combinations of two out of three. So that gives me either book and calculator or book and notepad or calculator and notepad. Actually, it's three. This is three by two divided by two. So this gives me three choices of the pair of these. Now, when I have already chosen two objects which are not pens, now I have to position them properly. Well, I have four objects. I can have four factorial different combinations, right? But I have two identical ones, which means I have to really divide it by two factorial because I can permute these two which are identical in any way without really changing the permutation, right? So the total number of permutations I should divide by the number of permutations of identical objects. That's part, by the way, of the lecture, one of the lectures which I did before. So that gives me, what? 12. So I have to multiply the number of pairs which I have chosen out of these three by the number of different positions four objects with two identical can take. So that gives me three times 12, which is 36. And this is, by the way, 24. So the total number is 72. That's the answer. It would be nice if you can come up with some other way to do it and get the same answer. All right. The next one. All right. So you have n points. In this case, n is equal to 6 n points. You have to pick k out of these points and make a polygon out of these k. Let's say I choose four points so I can make this polygon. Or I can have this polygon or any other. So the question is how many k-sided polygons can be formed from the n points. I'm not talking about convex polygons. I mean, whatever. Polygons can be, how should I put it? Something like this, this, this and this. This is polygon as well. Right? So any polygon and sides can even intersect. Now what's important is let's assume that there are no three points which are lying on the same line and no lines which connect these guys are intersecting in one point. So everything is completely random basically, right? So the question is how many k-sided polygons can be formed from n points. Okay, here is my suggestion. Let's do it this way. First, let's pick the k-points out of n. Because every other subset of k-points will give you a different polygon. So now we have different subsets. Okay, now, after you have a subset, you can make many different polygons. For instance, if you have let's say four points chosen, you can have this polygon, you can have this polygon and probably some others. Anyway, there are many ways to do the polygon. So basically what I'm saying is that if you will enumerate all these points, all these k-points which are forming the polygon, put them in the one sequence. This is number one, this is number two, this is number three, this is number k. And then you just draw from number one to number two, from two to three, from three to four, et cetera, to k and from k back to one, and that gives you a polygon, right? So presumably you have k-factorial different permutations of these k-points, how to put them in order and that gives you k-factorial different polygons. Yes, but there is one more consideration which we did not really take into account. If I have these four points, this is my number one, this is number two, this is number three, this is number four. So one, two, three, four, this is my order, okay? And that's the polygon, obviously, it's this one. But if I will put them in different order, two, three, four, one, two, three, four, one, it gives me exactly the same polygon. So what I'm saying is that any cyclical permutation, which means one goes to two, two goes to three, et cetera, or three, four, one, two, or four, one, two, three, any one of those gives exactly the same polygon. And how many of these cyclical permutations exist? Well, as many as many points exist, but I can start from any point and then go in sequence along the same polygon, right? So it looks like I have four. One more consideration. I can go in the reverse order, right? Something like four, three, two, one also gives me exactly the same thing, right? So I can go either from bigger to smaller or from smaller to bigger. Well, counterclockwise or clockwise, whatever it is. And so I have another K different permutations which are identical. So basically I have two K different permutations. It's cyclical permutations forward or backward, right? So that's why I have to divide it by two K. And that's the number of different polygons. So I divide it because all these permutations give exactly the same polygon. And that's the answer. That's the end of it. And well, I would suggest you to go back to the Unisor.com and try to solve all these problems again just by yourself. Now you basically know the solutions. If you can, try to come up with maybe a different way to solve the problem and get the same result. That would be the best actually. And in any case, I always encourage you to register on Unisor.com. It's a free site and it will give you an opportunity to basically take the course as a course, have some exams in between, etc. Just as an educational process, it's very, very useful. Thanks very much and good luck.