 In this video, I want to explore more the properties of binomial coefficients. Just as a reminder, in our lecture series Math 3120, we typically have the format that the first video in a lecture is about a new mathematical topic. The second video then usually focuses on logical methods, and the third video gives us some advice about mathematical communication. In lecture 18, because this is the last lecture on common torques, I actually switched the order of these, our logical topic came first because we'd explored some more examples of combinatorial proof that we had first learned about in lecture 17. But in this video, the previous video for lecture 18, we exclusively use combinatorial proof to study identities about binomial coefficients. In this video, I want to explore some more about binomial coefficients leading to Pascal's Triangle and the so-called binomial theorem. Let's review some of the properties, identities we already know about binomial coefficients. Let's take this first one, and this is the first one we found in our lecture series, that n choose k is equal to n factorial over k factorial times n minus 1 factorial, which that second one's in the denominator as well. This honestly was our definition of the binomial coefficients, and our lecture series, how we defined it. Remember where this thing come from? We were previously counting permutations for which we had these falling factorials, we had n fall by k, which was n factorial over n minus k factorial, and these falling factorials count permutations, which were ordered list of distinct elements. Now, if we want to move from ordered list to unordered list, we could forget the ordering, which then the division principle allowed us to divide the falling factorial by a k factorial, and then we forgot the order of the list. This makes an unordered list of distinct elements, which is what we call a combination, which is essentially just we're counting sets at that point, and that's where we got this formula, and we introduced the symbol n choose k to then count those combinations. But we're also discovering there's a lot more things about combinations, about binomial coefficients that we can learn about, like in the previous video we proved using commentarial proof, this identity right here, that n plus 1 choose k is equal to n choose k minus 1 plus n choose k. And this is one of our first examples in our lecture series of a recursive formula. So I want to say a moment about this idea of recursion. In the very next unit, starting with lecture 19 of this lecture series, we're going to start talking about integers, which will include topics like mathematical induction and recursion. If you haven't heard the term before, like if you've done some programming, computer programming, you probably are familiar with recursion. But recursion is an algorithm to calculation where you use the predecessors of a numerical sequence to then compute the successors in that sequence. So if you think of like the binomial coefficients as a collection of numbers, there's two parameters, two variables in play here, there's the number n and then there's also the number k there. So it's a double number sequence, but nonetheless it's a sequence of numbers. What this formula right here tells us is that we can compute the n plus one level binomial coefficients using the n level binomial coefficients. So if we know all of the binomial coefficients where the top number is n, we can then use them to compute all of the binomial coefficients where the top number is n plus one. And remember with sets, this top number tells you the size of the set, the bottom number tells you the size of the subset. If we can count all of the subsets of a set of size n, we can use that to count subsets of sets of size n plus one. But then also because there's a second parameter, we have to have some type of initial value. Well, actually this is true for any recursive sequence. We have some type of recursive relationship. It tells us how the next generation is affected by the previous generation, but we also have to have some initial value, something that gets us started. And with binomial coefficients, we also have discovered with our recursive relationships here, even without the recursion that if you take n choose zero, that is equal to one. Because n choose zero is the number of subsets of a set of cardality n that itself has zero elements, which there's only one such subset, it's the empty set, that's the only one. Now conversely, that's equal to n choose n, right? Which of course, that would say, we want all subsets of a set of size n that contains n elements. That would be the whole set itself. There's only one such set and hence these are both equal to one. Now also remember another identity we've seen previously is that if you take n choose k, that's the same thing as n choose n minus k. So choosing elements for a set is the same thing as choosing elements for its complement. And so with these identities in hand, we can put these together and actually work to compute some very interesting recursive formulas. So let's do some example of those right now. So using these identities you now see on the screen, we can do things like zero choose zero, that's equal to one by this property right here. We can do one choose zero, which is one. We also get one choose one, which is equal to one. Again, using these properties right here. We can do two choose zero, which is equal to one. Now we get to something like two choose one. How can we deal with that one? Well, that's where this recursion comes into play here. So by recursion, two choose one is the same thing as one choose zero plus one choose one, which is one plus one, which is equal to two. And then finally two choose two is equal to one. All right, let's do the next row here. Three choose zero is equal to one. Three choose one by recursion, this is the same thing as two choose zero plus two choose one for which two choose zero was one. We can see that two choose one was two, which gives us three. Next one, we're gonna get three choose two, which is actually equal to three choose one by this identity. So that's also three. And then lastly, we get three choose three, which is equal to one by our identities. Let's do one more example of this. Let's do the four row. Just to illustrate, I'll do a different color to make it stick out a little bit better because I'm afraid this is all blurring to us on the screen. If you take four choose zero, that's equal to one by the identity we've seen. You can take four choose one by recursion. This is the same thing as three choose zero plus three choose one, which is then equal to one plus three, which is equal to four. And one can actually prove in greater generality that if you take n choose one, this is always equal to n. Because of what we're doing right now, of course you can also use the standard formula that was on the top of the screen with the factorials as well. Four choose two, that's an interesting one. Four choose two is going to be, again, we can use the formulas using the factorials, but if you use this recursion here, this is three choose one plus three choose two, which we know all of these values. Three choose one is three. Three choose two is also three, so this is equal to six. The next one we have four choose three. We wear this as the same thing as four choose one, so that's equal to four. And then lastly, you get four choose four, which is equal to one as well. So be aware that I'm just showing you how recursion worked. I can use all of the previous numbers that I already know to help us compute the next numbers in the sequence there. And that's how recursion works. And the advantage of a recursive formula here is that if I know all of the previous information, I don't have to go through complicated calculations using factorials. I can actually compute simple sums, but it would be very useful if I can organize this information in a more better structure. And that's what's going to lead to what we call Pascal's Triangle, for which you can see Pascal's Triangle, at least the first couple rows illustrated right here. So I'm going to put my very first one at the very top of it, and this number represents zero choose zero, okay? And so then to keep things symmetric, the next row, which we had one choose zero and one choose one, that was one and one, we're going to get one and one, and we're going to put this in the midpoint right here. Then for the next row, we have the binomial coefficients with a two on top, you had two choose zero, you have two choose one, which was two, and then you have two choose two, which is one like so. So some of the observations we learned, when you take n to zero, that's always one, and if you take n choose n, that's always a one, every row of Pascal's Triangle is going to start and end with a one, okay? Another thing I want to mention is that every row of Pascal's Triangle is what we call a palindrome. It's read the same forward as it is backwards. So guess which way I'm reading it, left to right, one, four, six, four, one. You don't know because it's the exact same forward and backward, and that's because of this identity n choose k is the same thing as n choose n minus k. They talked about previously that there's a symmetric property with regard to the rows of Pascal's Triangle. And then another property that's really nice is the way that we've arranged it is the way that you compute a number, like say two, is you take the two numbers above it and add them together, one plus one is two, one plus three is four, three plus three is six, three plus one is four, six plus four is 10. So I can actually compute, when you arrange it as a triangle, you can actually compute the row involving the fives very, very quickly because you start with a one, you're gonna get one plus four, which is five, you're gonna get four plus six, which is 10, you're gonna get four plus six, which is 10, four plus one is five, and you end with a one like here. I can actually do the sixth throw pretty quickly as well. One, one plus five is six, five plus 10 is 15, 10 plus 10 is 20, and then the rest of it just repeats itself, 15, six, and one. I can do the seventh row, one, seven, we're gonna take six plus 15, which is 21, we're gonna take 35 plus 20, which is 35, and then at this point it repeats itself because I'm halfway through. And then I can very quickly compute these binomial coefficients using the recursion if I have Pascal's Triangle already constructed. Just again, a little bit of reference here I wanna make mention that each of the rows in Pascal's Triangle is marked off. This would be the nth row. It's marked off by the number on the top of the binomial coefficient, and you can actually always see that as the second number in the row. So this is the first row, the second row, the third row, the fourth row, the fifth row, the sixth row, the seventh row, like so. Then each, it's a triangle, so it's not a matrix, so there's not really columns, but we can count the positions. There's the zero position, the one position, the two position, the three position. There should be n plus one entries in the nth row, and that's your K value right there. So K equals zero. So this would be seven choose zero, seven choose one, seven choose two, seven choose three, seven choose four, seven choose five, seven choose six, and seven choose seven. So that's how you can read Pascal's Triangle. That's an important thing to be aware of, and it's just capturing all of these binomial coefficients. Using recursion, we can compute these binomial coefficients very, very quickly. And there are some beautiful properties that one can see in Pascal's Triangle. Like we said, you always start and end with a one. If you look at that slent diagonal, you always get the integers like so. That's kind of fun. Another interesting property is that if you take that slant, and then you just take the next position downward, that's actually the sum of that slant. One plus two plus three plus four is actually equal to 10. Similarly, one plus two plus three plus four plus five is equal to 15. Likewise, one plus two plus three plus four plus five plus six is equal to 21. So it turns out that if this slant gives you the, integers, the positive integers, then this slant gives you the triangle numbers, the sum of the integers, which we saw before, right? The triangle numbers one plus two plus three all the way up to N is equal to N times N plus one over two. Which when you look at that, it's like, you know, that does look like, doesn't that look like N plus one, choose two? Ah, so the triangle numbers are hidden inside of Pascal's Triangle, but you actually don't have to just do the triangle numbers. If you take the sum of any of these slants, like take this one right here, one plus four plus 10, this is not the sum of consecutive integers, but nonetheless the sum of those numbers is right there. One plus four plus 10 is 15 and it's not a coincidence. I wonder if there was an identity that we've already seen that could explain such a thing. And if there isn't one, maybe there wasn't one, could one prove this that the sum of these, some of these binomial coefficients always adds up to be this one right there? Could one formalize that and work from there? That's an exercise I'm gonna leave for the viewer here because my goal of introducing Pascal's Triangles is to introduce it to expose to these magical properties that binomial coefficients have. But I also have to finish this discussion of binomial coefficients with the titular topic. That is the binomial theorem. Because the binomial theorem is actually where binomial coefficients get their name. So if we don't give them the time of day, then we've done ourselves a great disservice. So the binomial theorem is interested in the following what we call a binomial polynomial. That is that we're interested in the binomial expansion of X plus A to the nth power. So X plus A is a binomial for which X is our variable, A is something else. It could be a number, I guess. You could have variables in it as well, it doesn't matter. But we have this binomial, we have two terms X plus A and we've raised it to a natural number power, so n. So it could be zero, one, two, three, four, five, et cetera. And so we're interested in what are the coefficients of the terms as you expand it? Because you go through the distributive law, the FOIL process, and it gets a little messy. Can we predict what those coefficients are gonna be without having to go through all of the nitty-gritty details? Now, again, I claimed n is a natural number, so we start with zero. n plus A to the zero power, that's just gonna give you one, anything to the nth powers is one. So you get the constant coefficient, has a coefficient of one, everything else has a coefficient of zero. If you take X plus A to the first power, that's just X plus A, so that means X will show up once, A will show up once, okay? That was pretty easy. X plus A squared, this is where we can use the FOIL method. So this is gonna be X plus A times X plus A, you're gonna get X squared, plus AX plus AX plus A squared. If you combine like terms, you end up with an X squared plus A, excuse me, two AX plus A squared. If you do X plus A cubed, what that, notice that by definition is X plus A times X plus A times X plus A, but since we already did X plus A squared, we can introduce that for the first two, X squared plus two AX plus A squared, and then you times that by another X plus A, what we can then do is multiply that thing out, for which case you're gonna get an X cubed, you'll get a, you'll get a, which one are we doing next? We're gonna do two AX times X, which is two AX squared. We'll do an A squared times X, which we can see that one right there. Likewise, you're gonna get A times X squared. We're gonna get A times two A, which will give us a two A squared X, and then lastly an A times A squared will give us an A cubed. Those are all the things, those are all the six terms you get there. Combining like terms, there should be two X squares, there should be two Xs, and then so that combines to give us X squared plus three AX squared plus three A squared X plus A cubed, huh. Those numbers, if we look at just the coefficients, they look a little bit familiar to me. Let's do one more example. If you take X plus A to the fourth power, I claim that if you multiply it out and combine like terms, your coefficients will look like X to the fourth plus four AX cubed plus six A squared X squared plus four A cubed X plus A to the fourth. And the coefficients there are gonna be one, four, six, four, one, one, three, three, one, one, two, one, one, one, one. All of those numbers look very, very similar. And if we were to bring back up the screen, I'm actually gonna zoom it out so we can see them together. If you look at the coefficients of our binomial expansion, they exactly look like the numbers in Pascal's triangle. One, one, one, one, two, one, one, three, three, one, one, four, six, four, one. If you didn't pause the video, I would actually recommend that you do so you can check out this calculation yourself. And even if you're really bold, try X plus A to the fifth power. And I want you to convince yourself that that thing is gonna be X to the fifth plus five AX to the fourth plus 10 A squared X cubed plus 10 A cubed X squared plus five A to the fourth X plus A to the fifth right there. That's actually gonna be the value it simplifies into at the very end. Kind of interesting. They're the coefficients from Pascal's triangle. So these are the binomial coefficients. And how does that work? Well, when you look at these expressions here, notice that when you look at the different monomials, basically the monomials, you're always taking some collection of A's and some collection of X's so that the power, when you combine them together, adds up to be the power you're working on. So you either get the fourth power or the fifth power or whatever, depending on what you're looking at there. And how do you produce something like A squared X cubed? Or how do you get something like A cubed X squared? Well, that basically comes down to you have like an A, A, A, X, X. That's a possibility. You could do A, A, X, X, A. You could do something like A, X, A, X, A. You have these lists, these binary list of terms for which you either choose an X or you choose an A. The length of the list has to be five, but there's binary list. And so you could sort of ask yourself, well, okay, I wanna look for all of the list which have three A's in them. Well, how many ways can you make a binary list with exactly three A's if there are five letters, five length total there? Well, basically you have to choose where the A's are gonna live. They can live in the first, second, and third position. They can live in the first, second, and fifth position. They can live in the first, third, and fifth position, and all the other possibilities. You have to choose which buckets are gonna hold the A's and then by default, which buckets will hold the X's. And if I have five buckets and three A's, I have to position, how many ways can you do that? That's going to be five, choose three, is it not? And so the number of combinations you can create here is in fact, these binomial coefficients. Let's formalize what we've been saying here and then ends with this so-called binomial theorem. The binomial theorem tells us that the expansion of the binomial X plus A to the power N is gonna equal X to the N times N choose zero, A X to the N minus one times N choose one, A squared times X to the N minus two times N choose two. And then all the way through, if you look at the most generic term in this sequence, you're gonna get N choose K. And this will then look like N choose K times A to the K times X to the N minus K. And that's what each and every one of these terms looks like as you work through it. You always have some combination of X's and A's so that their powers add up to be N. And then there's a kind of a binomial coefficient which the binomial coefficient is the coefficient of the A, how many times you chose the A. And so a more compact form, you get this right here. And so I do have the proof formally written out here, but honestly we've gone through all of the details of it already. With the binomial expansion, you have X plus A to the Nth power. This will be X plus A times X plus A times X plus A. And so we're interested in what is the coefficient of A to the K times X to the N minus K. Now it's very possible the coefficient could be zero because if that monomial is not possible to create, you could get a zero right there. That's a possibility. Now, how do you form a C? Well, you're gonna get, C is gonna be the sum is gonna be formed. The coefficients be summed because you add together all the like terms. You have to grab all of the A Ks, X to the N minus Ks that show up in the product. And how many of those you get is what C is gonna be. Well, how many such terms are there? Well, to produce such a monomial A to the K, X to the N minus K, you're looking for words, like, because of the products as you foil out the X plus A times X plus A times X plus A, you're gonna get some words, some combinations of X's and A's like we were considering beforehand. Because the multiplication is commutative, I can always move all of the A's to the front and all of the A's to the end. And so that's how you produce these monomials. So we have to count how many binary words, how many binary sequences can we form with K A's and N minus K X's? But since in the end, we just, we're just gonna put them back, right? All of the A's to the front, all of the X's in the bottom. It really would just have to count that which positions were the A's in originally? And that comes down to these binomial coefficients that we already chose. So the number of such words is gonna be in choose K. And therefore the coefficient is going to be in choose K as well. And so that then gives us the binomial formula which we had up here that the, sorry, there's some K's and J's mixing up here that all of those J's should be K's inside of our formula. The binomial expansion comes to these binomial coefficients. And this is why we in fact call them binomial coefficients because they're the coefficients of this binomial expansion. Even though we can use them to count so many other things, it's a very, very important tool for the binomial theorem. Let's just do a quick example of this real quick. So let's use the binomial theorem to expand X plus two to the fifth power. So what that's gonna look like is we end up with a five choose zero X to the fifth power plus we're gonna get five choose one times two to the first X to the fourth. We're gonna get a five choose two X squared times X cubed, sorry, two squared times X cubed. Then we get five choose three which is gonna be two cubed X squared. We then get five choose four times two to the fourth X. And then lastly, we're gonna get five choose five. We then get two to the fifth and X to the zero. So I'm not even included at that time. And so then as we evaluate these things, these binomial coefficients, we can read off the fifth row of Pascal's triangle which gives us one X to the fifth. We're then going to get five times two to the first which is two X to the fourth. Then we're gonna get five choose two which is 10, two squared is four, we get X cubed. Then the next one, we're gonna get five choose three which is also 10, two cubed is eight, you get X squared. Then five choose four is five, you're gonna get two to the fourth is 16, X. And then lastly, you're gonna get five choose five which is one times two to the fifth is just 32. And now if we simplify those coefficients one more time, we end up with X to the fifth plus five times two is 10, X to the fourth, add to that four times 10 which is 40, X to the third. Then we're gonna get eight times 10 which is 80, X squared. Then we're gonna get five times 16 which is also 80, 80 times X. And then lastly 32. And so this very last line written in blue is then the binomial expansion there for which we did it very, very quickly using the binomial theorem assuming that we can compute all of the binomial coefficients. But if your exponent's not really big you can actually compute all of these binomial coefficients very quickly from Pascal's triangle because you need the whole row. And this is an example where the recursion actually makes the counting a lot, a lot easier. And so this notion of recursion is gonna help us lead into our next unit of integers that we're gonna start considering in lecture 19.