 Hi, I'm Zor. Welcome to Unizor Education. This is the fifth session of combinatorics problems. You might have noticed that I really like these problems because the purpose of the whole course is well to train your mind to to develop your abilities to solve different problems and combinatorics problems are really very much in that particular domain of well developed tools to basically develop your analytical abilities. All right, so I also hope that you did go to Unizor.com first read through these problems and try to solve them just by yourself. There are answers provided as well as well as explanation of the logic, but the logic I will basically explain here. Okay, so without further ado, let's me just address one problem after another. So problem number one is you have N different lines, straight lines on the plane. Well, they divide the plane into some parts, some areas. Like in this particular case, I have three lines and they divide into one, two, three, four, five, six, seven different parts. They divide my plane. Well, the question is what's the maximum number of divisions, maximum number of parts and line on the plane can divide that particular plane? Well, when I was trying to solve this problem, my first attempt was, okay, let me just try to, you know, have some experiments. Well, if I have only one line, it divides the plane in two parts, right? So that's kind of obvious. If I have two lines, well, if they are parallel, they divide the plane into three parts, but we are talking about maximum, which means they have to cross, they have to intersect, in which case they divide in four parts. Now, three lines, I was just drawing a triangle of three lines and they divide into seven parts. Well, let me try the four lines. Well, and they should all intersect each other, right? How can I make it one, two, three, four, something like this, right? So let's count one, two, three, four, five, six, seven, eight, nine, ten, eleven, eleven parts. So four gives me eleven. I would like to find some kind of law, basically. And quite frankly, if you look at this, it's actually quite obvious. Because this is plus two, this is plus three, this is plus four, right? From two to four, I have to add two. From four to seven, I have to add three. So the third line adds the three. The second line adds the two. The fourth line adds the fourth, right? So that looks like the law, right? So it looks like if I would like to draw the N plus first line, if N has already been drawn, and I'm drawing N plus first line, and the number of parts, which this all picture of all these lines together divide the plane is some kind of function of the number, then the previous number should be added the number which this line actually has. So the second line adds two to whatever was before. Now, before it was two, right? One line divides in two parts. So my second line, if you add two, you will be four. Now, my third line would add three to whatever it was before, and before it was two lines and four parts, so plus three, seven, etc. So it looks like this, but this is not a proof, obviously, right? So I kind of guessed the formula, the recursive formula, which basically indicates the law. And by the way, I have an initial condition f of one is equal to two. So the number of parts, my one line separates the plane, the number of parts is equal to two. So basically using this and my absolutely not proved yet, just to guess the formula, I can come up with any number like from two, I get to four, from four to seven, from seven to eleven, etc. But I have to prove this formula somehow. Or, well, the proof in this particular case might be a little strong word. At least explain. Alright, so let's just think about explanation. What's behind this formula, actually? Think about it this way. Let's consider you have certain number of lines already drawn on the plane. And what I'm going to do is I'm taking one particular part, let's say this part, and take two points which are on the border of this part and connect them with a straight line, which does not intersect anything else. So I'm talking about only the segment of the line which is attached on both ends to the edges of some part. Now, if I draw this particular segment, it has no intersections in between these two ends. Then what I'm basically doing, I'm adding one to the number of parts, right? So if I had whatever number of parts before, I draw this line which basically divides my one particular part into two pieces. So I'm adding one part to whatever was before. It's one part more. What's important is that this line doesn't intersect anything else. So it's only in between these two edges of this particular segment where this particular line is inside the part and no other intersection. Same thing anywhere else. If I were to take, let's say, this particular triangle in the middle, if I'm taking two points here, connect them with a straight line, what do I do? Basically I divide one part into two, which means I'm adding one particular part to the total number. Okay. That being said, let's consider a situation when you already have n lines and you are drawing the next line, the fourth line. I mean, in this particular case, the fourth. Generally speaking, n plus first line. Now, the n plus first line intersects with all other n. Now, intuitively it's kind of obvious that the more intersections, the more parts are added and we are looking for the maximum number of parts, which means my new line, if I would actually like to add as much as possible to the number of parts, it should intersect all other or all old lines. Now, there were n old lines, so I will have n intersections, right? Now, n intersections actually means that I have n plus one segments. I have three intersections and four segments. Now, and I was just explaining that every segment actually adds one more part to the total number of parts. So, this segment adds one more part because it divides this big area into two. Now, this segment divides this area into two and this segment divides into two and this one. So, every segment divides some part into two, which means that the new number of parts is exactly equal to the number of segments. And the number of segments which one line intersecting with n other is divided is actually n plus one, right? So, n division point, so we have n plus one segments. And that would actually justify this particular equation. So, I have to take whatever the number of parts was before and I have to add n plus one, which is number of intersections the n plus first line has with all old lines. Alright. So, basically I consider this, well, again, not maybe as a proof. Proof is probably a stronger word. I can obviously do exactly the same thing by induction, but the logic is exactly the same. So, I consider this formula as fully explained, if not fully proved. Okay. Now, based on these two formulas, can I come up with just an expression, an algebraic expression of what is the maximum number of parts which n lines divides the plane? Well, yes, I mean, obviously we can. Just think about it this way. Let's just forget one for a second. What is f and n plus one? Well, if n is equal to one, then I have to add one, right? If n is equal to two, I have to add two. So, it's basically sum of n numbers. That's what f plus n is, right? So, if I would like to know what is f at one, I put just one and that's one. If I want to know what's f at two, I have to add two of them, one plus two, which is equal to three. So, you see, this is n, this is n. So, if I don't have this one, my f at n actually represents the sum of numbers from one to n. Well, plus one, I will add one to the formula, actually. Now, what's the formula for this f at n? Well, I mean, if you don't remember it, and I never remember any formulas, I just do whatever the typical is for arithmetic progression. I reverse the sequence and add them together vertically. I have two f at n is equal to n plus one, two and n minus one, again, n plus one. So, I have n plus one repeated n times, right? So, f at n is equal to n, no, n plus one over two. Well, that's almost the formula, except I have to add one. So, my final formula is, or at least it seems to be, now, is it true? Let's just check it out. f at one, this is one, this is one, this is two, divide by two is one plus one, it's two. Correct. Now, is this particular rule observed? Well, let's check it out. f at n plus one equals, if I will substitute n plus one here, what will I have? Instead of n, n plus one. Now, how can I simplify it? Well, let's do it this way. n plus two, I will actually separate n and two and multiply by n plus one, because I have to have n times n plus one, right? So, it's n times n plus one, plus two times n plus one, divided by two, plus one, right? Now, I will divide each separate member of this sum, and I will get n, n plus one, over two, plus two, divided by two, it will be just n plus one, and one. Now, let's group them together. Now, what is this and this? Well, that's f at n, right? So, I have that f at n plus one is equal to f at n, which is this and this, plus n plus one, which means exactly what we have to receive. So, my main recursive rule is observed, and that's why that ends the formula, basically, the proof. So, that's the answer and the explanation. Now, what was important in this particular case? I think the most important part of this was to realize that from the geometric standpoint, if you have certain division of the part of the plane into different parts, any segment which connects two points on the border of one particular part without intersecting anything in between, add one to the number of parts. That's the most important part. That's the most important logical consideration because as soon as you find out that this is basically the principle when these parts are formed, then everything else is easy because it's obvious that the n plus first line should cross n previous lines to get the maximum number. So, to convert this into segments, n plus one segments, and that's the number which you are adding every time you're adding a new line, that's the piece of it which is important. Okay, the second problem is very similar, but instead of straight lines, we have circles. Now, circles are slightly different, but again, one circle gives you two parts inside and outside. Two circles gives you one, two, three, four parts. Three circles, oh, that's difficult, right? One is outside, two, three, four, five, six, seven, eight, seems to be eight. And the four circles, well, I'll try to draw four circles and then corresponding number of... So, we have to intersect all three circles before that. So, how can I make it? It's something like this. So, it intersects each out of these three, right? So, that's something like this. Well, let's count. Let's count. One is outside of everything. Now, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve, thirteen. Is that it? Thirteen. It should be something else, unless I didn't cross it correctly. Oh, yes, fourteen here. I remember it should be fourteen. Okay, that's enough. Enough preliminary analysis. Now, let's basically think about what's the law. Well, again, let me try to check what's the differences. This is plus two, this is plus four, this is plus six. So, it looks like I'm adding even number every time. So, it looks like effort n plus one is equal to effort n plus... If this is two, I add two. If this is three, I add four. If this is four, I add six. So, it's two n, right? So, the previous number is doubled. This is n plus one, n is a previous number, so it's doubled. For two, I take one and double. For three, I take two and double. For four, I have three and double. So, that's how it looks. With the first one, if I have only one circle, it breaks the plane into two parts. So, that seems to be my formula. The question is, let's just think about how to logically explain why do we have this particular thing. Now, if you remember with the straight line, I was talking that the number of segments actually is the additional number of parts. Now, why? Because I had actually n previously drawn lines, and I'm drawing n plus first line. So, I have n intersections and n plus one segments. What do I have here? Well, obviously, I should really consider again the case when one new circle, n plus first circle, is intersecting with all n previous sectors, right? But here is the difference. Line and line intersect in one point. A circle in a circle, generally speaking, I mean the most number of points. That's two, right? Two circles, if they intersect, it's always two points. Which means that the number of segments, number of new segments, which every circle brings into the equation, is different. Because if I had n circles before, I have two n intersections, right? With each circle, I intersect twice. My new circle, n plus first circle, intersects twice with each previous one. So, the number of new points is two n. Now, on a circle, number of intersections is exactly the same as number of arcs, right? So, instead of segments for the straight lines, we're using arcs for the circles. But the principle is exactly the same. I mean, if you have certain area and you add an arc or whatever else, which connects two points on the boundaries of this particular area, then you add a new part. So, every arc adds the part. Number of arcs for the n plus first circle into which the intersection points divided is two n, right? N circles, old circles, so it's two n new intersections. And that's why this explanation actually is valid. So, now with these two in mind, I actually have to construct the formula which does exactly this. Now, explanation is exactly the same as before when I was talking about arithmetic progression. This is also kind of arithmetic progression with different difference. And the formula actually is n times n minus one plus two. It's very easy to come up with this formula using just a recursive law. But let me just prove it by induction. If n is equal to one, I do have two. Now, if I will take n plus first, it would be this, right? So, instead of n, I put n plus one. So, instead of n would be n plus one, n minus one would be n. But if I will change it to n times n minus one plus two, n times square. Now, let me just do it more accurately. So, this is n square minus n plus two, right? Now, this is n square plus n plus two, which is equal to n square minus n plus two plus two n, right? Minus n and plus two n. But this is f of n plus two n. So, we get exactly this word. So, this particular, so this is f of n plus one. So, this particular law, recursive law is observed. Well, that's it for this particular point. Now, I have two simple problems. Solve the equation. Well, it should not really scare you. I mean, I'll just open it up. What is number of combinations from four by x? So, it's four factorial divided by four minus x factorial and x factorial, right? So, I reverse it since it's one over, so I'll have this one. Now, similarly to this, I will have five minus x factorial, x factorial, five factorial equals six minus x factorial, x factorial and six factorial, right? Now, x is supposed to be an integer number, obviously, because we're talking about number of some objects, right? And it's positive integer number. So, the factorial is definitely positive, so I can just reduce all this. Now, if I can also reduce by four minus x factorial, because again, it's an integer number. Now, what is five minus x factorial? This number is one greater than this one. So, the last one, five minus x, remains. But before that, it was four minus x, three minus x, etc., they will be divided by four minus x. And in this case, I will have two less numbers in the factorial. The previous five minus six and this one, six minus six, because starting from four minus six, I'll reduce by four minus x factorial. So, I have one over four factorial minus five minus x divided by five factorial equals five minus x, six minus x divided by six factorial. Now, I will multiply everything by six factorial. I will have this. Now, six factorial and five factorial, that would be six here, right? And this would be six factorial divided by four factorial is five by six, it's thirty. So, that's my equation, it's just plain quadratic equation. So, it's thirty minus thirty plus six x, so that's what it is here. Thirty minus eleven x plus x square. So, x square minus seventeen x plus thirty and the roots equals to zero. The solutions are fifteen and two, because they're integer solution, they're very easy to basically guess. The sum is equal to seventeen, which is plus seventeen here and product is thirty. Now, obviously, fifteen is not good because it's too big a number for, it's supposed to be from zero to four, right? Because it's a combination. So, the two is a solution and you can check it out. I skip the checking here, I do put it under notes to this lecture. And the last problem is, consider a point on the plane and end different rays from it. How many different acute angles they form? Well, this isn't easy thing because any two rays, this one is one angle, this one is another angle. So, any two rays form one acute and one obtuse angle. Well, let's just consider that rays do not really lie in the same line, just to be more general. So, basically the solution is c of n, number of combinations of two objects from n. Basically, any two rays gives you the angle. Well, that concludes this particular lecture. Please go back to Unisort.com and review all these problems yourself. That's very, very helpful. And good luck!