 Well, welcome to episode 36 on the binomial theorem. Today we're going to be looking at the expansion of binomials, powers of binomials, to read the large powers, and how we can compute them quickly. Let's go to our list of objectives for this episode. First of all, we're going to be looking at patterns that we find in the expansion of binomials. What I mean by that is if you were expanding like a plus b squared, a plus b cubed, a plus b to the fourth power, how we could find patterns in the coefficients. And then we'll look at something that we call Pascal's triangle. This is named for a French mathematician named Blaise Pascal. Then we'll move on to factorials and what we call binomial coefficients. And then finally, we'll look at a theorem that was first proved by Isaac Newton back in the around 1700, just before 1700, called the binomial theorem. Okay, well, let's begin by looking at patterns in binomials and the expansion of binomials. And let's go to the next graphic and I think you'll see what we mean. Okay, first of all, we have a binomial, a plus b, and we're going to raise it to the zero power. So what is a plus b to the zero power? One is one. A plus b isn't zero. That's exactly right. If a plus b is not zero, because zero to the zero power is undefined. That's a good point, Stephen. So this answer is one otherwise. And whoops, let's see. I must better go to a different marker for that. So we get a one there. Yeah, so in all of these, I'll be assuming a plus b isn't zero. Now, in the next case, a plus b to the first power is a plus b. So I'll write a plus b right here. You notice I'm kind of spreading it out. You'll see why in a moment. What is a plus b squared? A squared. Okay, a squared. Plus two a b. Plus two a b. Plus b squared. Plus b squared, yeah. And now finally, a plus b cubed. That's not necessarily a formula people generally remember. Does anybody know what a plus b cubed is right off hand? A cubed. Okay, a cubed. Plus three a squared b. You're right. Three a squared b. Plus three a b squared. Three a b squared. Right, very good. Plus b cubed. Plus b cubed. Okay, now what I wanna do is look at some patterns that we see in these expansions of binomials. First of all, you notice that the number of terms, like here I have three terms, is one more than the exponent. If there's a square there, I get three terms. If there's a cube, I get four terms. It even worked up here. When there's a zero, I get one term, namely, namely one. So we always have one more term than we have for the power. The next thing we might notice is the powers on a go down by one. I have a cube, a squared, a to the first power, and then essentially a to the zero. They're gone, a to the zero is one. At the same time the powers on a are going down, the powers on b are going up, I have b to the zero here, or one, b to the first power, b to the second, b to the third. So if the a's are going down and the b's are going up, what that means is I have the same number of variables in every term. So here I have a three times. Here I have two a's and a b's, that's a total of three. So this is referred to as a term of degree three. This is a term of degree three. I have an a and two b's. And this is a term of degree three. All the terms are of degree three. So if in the future, if you're expanding a binomial to some larger power, one of the things you want to notice is every term should be of that same degree as the initial power, the cube, right here. So what portion of this would be the really awkward part to find then? The coefficient. It would be the coefficients, right. So what we need to do is turn our attention to the coefficients and see how we could find the coefficients, like in this case, the coefficients are one, three, three, and one. Now you know, the way we can find the coefficients, say for this a plus b cubed, is what if I go back up here to a plus b squared and multiply it by a plus b? By a plus b. Now, for example, to get the a cubed, I just multiply a squared times a, and that gives me a cubed. But I think there are several ways I can get terms that say a squared b. Let's see, here's an a squared. If I multiply that times b, so if I multiply a squared times b, I'll come up with one a squared b. But what's another way I can come up with some a square b's? Two a b times a. Two a b times a, exactly right. I can multiply these two terms together, and I'll get a squared b's. In that case, I'll get two of them. And so I get one, let me just put a one on top of that, I get one from this product, and I get two from this product. Now, one plus two, that makes the total of three that we have right here. And you know, since I'm multiplying by one a and one b, the coefficients one and two are the same as the coefficients that I see right here, one a squared and two a b. So I'm thinking there's a pattern in this, and that is that if I add the two coefficients that I see right here, I'll get the coefficient that I need right below it. Let's just try the three right here. If you look at the two coefficients right up above it, there's a two and a one. And if I add the two and the one, I get a three. And you might say, well, Dennis, what's the connection? So just because that's a two and a one, why should this be a three? Well, let's just remove these arrows, and let's see if we can verify this pattern for this other term. Here we have a b squares. Now, how can I get a b squares? Well, I can take the two a b times b, and I'll get a b squared. I'll get two of them, which is exactly what that coefficient is. I get two of these. And I can also get some a b squares right here with this product, and I'll get one of those. And that's exactly the coefficients that's on the b squared. You see, this coefficient won't change because I'm just multiplying by one a or by one b. So that won't affect the coefficient. So if I take the two and the one and add them together, I'll get a three right here. That even works on the very ends because the only way I can get an a cube is to take one a squared times one a, and that gives me a cube, or one b, and one b squared gives me b cube. So the pattern is that if I add the two coefficients directly above, I'll get the coefficient on the term just below it. Now, this leads us to what we call Puskal's Triangle. And let's go to the next graphic and we'll see this. Blaise Puskal was a French mathematician who lived in the early 17th century. First half of the 17th century, obviously in France, I guess, Blaise Puskal, and his triangle begins with a one at the top, and then two ones down below. And then if I continue the ones down the diagonals, ones down the diagonals, the numbers in the middle are merely the sum of the two numbers directly above it. One and one is two. And the same thing here, one and two is three. And two and one is three. And on the next row, we'll see the ones are coming down on the ends. One and three is four. Three and three is six. Three and one is four. And this pattern continues, so actually there's really no end to Puskal's Triangle. What would be the next row of the triangle? What number would go right here? One. One, okay, and then here? Five. Five. Okay, and then? Ten. Ten. Ten. Five. Five. One. And one. Now with those numbers, we can now expand A plus B to the fifth power. These are the coefficients of the terms. And now I know it's the fifth power because if you look in the second diagonal, one, two, three, four, five, that's the power on the exponent. These were the coefficients for A plus B to the first power. Here we have the coefficients for A plus B squared. A plus B cubed. A plus B to the fourth, and now here's the fifth power. You notice there are six numbers written across there, and we said that if you expand A plus B to a power, you get one more term, you get one more term than you get the power. So a fifth power, I should get six terms. Okay, now I'm gonna write those numbers down. A, one, five, ten, ten, five, one, and expand A plus B to the fifth. In fact, maybe there's enough room right below this problem to write this out. So I think you'll probably have to remove the box over on the lower right for us to squeeze this in. But A plus, oops, A plus B to the fifth power is going to be, first of all, one, A to the fifth. The next term is gonna be A to the fourth B. And what's the coefficient? Five. A five, okay. And the next term is gonna be A cubed B squared. You notice the As are going down, B's are going up, the coefficient is? Ten. Ten. And the next term will be ten A squared B cubed. And the next term would be five A, B to the fourth. And then the last term is B to the fifth. So by writing out Pascal's Triangle, I have the coefficients for the expansion of the binomial. And that's really the only subtle part because everything else is pretty straightforward. The As go down, the B's go up. Every term has degree five. Degree five here, degree five here, degree five here, et cetera, all the way across. Okay, looking at Pascal's Triangle, there are some other interesting patterns that we can find in the Triangle. You notice coming down here, we have ones on these two diagonals. What about this diagonal? What did we say? Consecutive integers. They're just consecutive integers and they happen to be the exponents of the A plus B that's being expanded. What about the numbers we see here? One, three, and six. One, three, and six. Have you ever seen numbers like that before? The next number was gonna be ten coming down that way. Also one, three, six, ten coming down this way. These numbers are called triangular numbers. Let me just show you up above. Here's why they're called triangular. One, I could represent with one dot. Three, I could represent with three dots. Six, I could represent with six dots like this. In other words, I could make triangles out of them. Could I represent ten as a triangle? Sure, I'd start with four dots. One, two, three, four. One, two, three, one, two, one. In what sport shall we say do you see an arrangement like that? Pool? Yeah, pool, except pool has 15 balls and I'm thinking of just regular, what they call straight pool, 15 balls. So when you rack the balls, it's one through five and there are 15 balls. But where do you see these 10? Well, this is something I used to do a lot when I was in high school, let's go bowling. So 10 dots, you see the arrangement of the bowling pins that way. Okay, so those are called triangular numbers and along the diagonals we see those triangular numbers. Another pattern is look at the sum along every row. Well, the sum on this row is one. The sum on this row is two. The sum on this row is four. What's the sum on this row? Eight. Eight, okay. And the sum on, let's say I'll have to go to the very end. What's the sum on, I'm gonna call this row four because there's a four there. What's the sum on row four? 16. 16. Let's say we have one, two, four, eight, 16. What pattern do you see there? Powers of two. Those are powers of two, yeah. And I've erased row five, but those would add up to be 32. So in other words, we could hazard a guess that if you expand an expression like A plus B to the nth power, that the sum of the coefficients in A plus B to the nth power is two to the nth. You see this is two to the fourth, and this is two to the third, this is two to the second, and this is two to the first, and this one is two to the zero right here. So imagine there's a zero right there. So if you're expanding A plus B to the nth, then all the coefficients should add up to be two raised to that power. So yet another pattern that we find in Pascal's Triangle, and there are still other patterns in it, but we don't need to get into all that. Okay, well rather than just talk about it, let's actually use it. So let's go to the next graphic and see how we could use Pascal's Triangle to expand a few terms. I'm gonna work these out on the green board, but the first problem says expand A plus B to the fifth. You know, I think we've actually done that one already, so I think I'll skip that, A plus B to the fifth. But then we wanna find X plus three to the sixth, five Y minus one to the fourth, and now here's a complicated one, U cubed minus 10 V squared cubed. Now I say it's complicated, but actually it's no more complicated than any of these others, and they're all fairly easy to work out. So I'm gonna go to the green screen and work these individually, except for that one, because we've already done A plus B to the fifth, and let's see how we would expand every one of these using Pascal's Triangle. Okay, so if we come to the green board, I'm gonna write up here in the corner the numbers that I see in Pascal's Triangle. One, two, one was that row. What was the next row? Could you tell me what that was? One, three, three, one. One, three, three, one. Yeah, and the next row was... One, four, six, four, one. One, four, six, four, one. Yeah, and then I think that's as far as we had printed on the graphic, but then we found one more row a little bit earlier. And of course, this just keeps on going over here. Now, the first problem that we had was to find X plus three to the sixth power. So let's do that right here. X plus three to the sixth power. So, do I have enough rows in my triangle to work this problem? Not quite yet. I don't think so, because I need to get to row number six. That's the one in the... If you look at the second diagonal, I need to go to row six to go with that six. Now, that means I'm calling this one row zero up here, and then row one, two, three, four, five, six. So let's fill in row six. Who wants to tell us what row six will be? Susan, could you tell us what row six is? One, six, 15, 20, 15, six. And one, right. You know, one of the patterns that we have in Pascal's Triangle, I should have mentioned, it's fairly obvious, is that the triangle is perfectly symmetrical. So if you go from the middle out, 15, six, one, and here 15, six, one. So if you go right down the middle, if you could draw a line right down the middle, of course it would split it on the row five, then it's perfectly symmetrical. So if you know the first half, you also know the second half of the triangle. Okay, now when I expand this, I'm gonna get, well, I'm gonna treat X as if it were A, and I'm gonna treat three as if it were B. So this would be, actually maybe I better go down here to make sure I have plenty of room. This would be one, X to the sixth, plus six, X to the fifth, times three to the first. See, that's my A, B term. A to the fifth, B to the first. And then I'll have 15 X to the fourth times what? Oh, nine, okay, I'm gonna write it as three squared just so we see that that's a sixth degree term where I'm thinking of this as my B. And then I'll have 20, 20 X to the third times three cubed. And then I'll have 15 X squared times three to the fourth. And then I'll have six X times three to the fifth. Plus, I'll have to squeeze this one in underneath, plus one times three to the sixth. Now you might say, Dennis, this is terribly long and complicated. Well, what's the alternative to writing it out this way? What's the alternative to finding X plus three to the sixth? Multiply it out. Multiply it out, yeah, you'd have to write this down. X plus three times X plus three times X plus three and you see what's gonna happen here. And multiply it out, multiply that out by regular multiplication. Now that's what takes a long time. This is fairly simple child's play. Okay, so if I go back and simplify this, I have X to the sixth plus 18 X to the fifth plus nine times 15. What's 10 times 15? 150. Okay, so what's nine times 15? 135. Yeah, just be 15 less. That's gonna be 135 X to the fourth plus 20 times, what's three cubed? 27. Yeah, 20 times 27. It's gonna end in a zero. What's two times 27? 54. 54, that'll be 540 X cubed. Okay, so we're doing pretty good here. Now three to the fourth is 81, 81 times 15. Oh, 81 times 15, let's see if we can do that without a calculator here. If I wanna find 81 times 15, I'll tell you what, what is 80 times 15? 80, 15. Well, wouldn't that be the same thing as 30 times 40? Yeah, what I did was I doubled that number and I cut this one in half. 30 times 40 is 1200, so this is 1200. Okay, but that's 80, 15s. What's 81, 15s? 1200, 15? 1215, yeah. Isn't that the year the Magna Carta was signed, I think? Okay, 1215. Just had to see if you were paying attention there. Okay, 1215 X squared plus, now let's see, 81, that was three to the fourth. What's three to the fifth? Well, I have to put another three in there. To 243. 243, 243 times six, I think I've reached the point where I'm gonna have to go to regular multiplication. 243 times six is eight, five, two, six. Did I do that right? I think I did. So two, six, five, eight times X, plus, okay, now we come to three to the sixth. Oh my gosh, three to the sixth. What's three times this number? That's three to the fifth. I think it's 729, plus 729 right here. Okay, so this is the expansion of X plus three to the sixth power. Did you see a mistake? Yeah, 243 times six is 1458. Oh yeah, of course, yes. Thank you. You know, I'm better off using my shortcuts than in regular multiplication. Yeah, that's gonna be 18, carry one, 25. This is 12, so not 24. So that's gonna be 1458. What happened in 1428? 1458, can anyone tell us? Good, I can't either. Okay, 1458 and 729. Okay, so I think now we have the coefficients right. You know, what was the hardest part of this problem? It was the multiplication. That's where I made a mistake. The actual expansion is pretty simple, actually. If you know Pascal's Triangle. Now you might say, Dennis, when was the last time I ever had to expand a binomial to the sixth power? You may be asking yourself that. When are you gonna have to expand X plus three to the sixth power or something comparable to it? Well, you just never know, but I think what's important is that you become familiar with the patterns in the triangle because these things come up, well, in other courses now and then, and you'll see our references to Pascal's Triangle. Let's go back to that graphic that had the examples on it and let's pick another problem from those examples to work. Yeah, we've just done part B. Let's go to part C, five Y minus one to the fourth. Now, here are some of the differences that we see in this problem. It's a difference rather than a sum. And the first term is not a simple A or X, but it's a five Y, but we do have a rather simple second term raised to the fourth power. Okay, so here we go. Let's try working out. If you go back to the green board, let's try, oops. You know what, I just erased Pascal's Triangle. Let me just fill it in right here. Somebody help me out. What's the next row? One, two, one. One, two, one. Thanks a lot. What's the next row? One, three, three, one. One, three, three, one. The next row. One, four, six, four, one. Oh, that's cheating. He's reading it all off his paper. I kept mine. Lene, what's the next row going to be? One, five, 10. 10. 10. Five, one. Five, one. Very good. Okay, and we could keep on going as far as necessary, but I think this is enough. We wanted to find five Y minus one to the fourth power. Yeah. Well, let's see, how are we going to handle that negative? I think we could write this as a sum if I write it this way. Five Y plus what? Negative one. Plus a negative one, okay? So what I'm going to do is just treat A as if it were five Y, and I'm going to choose B as if it were negative one, and let's expand it. The very first term is going to be one times five Y raised to the fourth power. Now that's everything there, both the five and the Y are raised to the fourth power. Plus, what's the next coefficient going to be? Four times five Y to the third, right? Times negative one. Exactly, very good. You see, we're using this row because this is a fourth power, the one, four, six, four, one. So that's where we get the four from right here. Five Y cubed, negative one to the first. Then the next term will be six times five Y squared times negative one squared. And then the next term will be what? Anybody, somebody other than Stephen, he answered several of these others. What's the next term going to be? What's the coefficient? Four. Four, yeah. What else, Susan? Five Y. Five Y. Times negative one to the third. Times negative one to the third, and I see I keep crowding myself in on the last term, and the last term is going to be plus one times negative one to the fourth. Okay, so what I do is I write out these terms in factored form, then I go back and reduce them. And let's see, five to the fourth. That would be 25 squared. Does anybody know what 25 squared is? Is it 625? He's asking me, is it 625? It is, it is, 625. You know, the way I do that is I think if I had 25 quarters, I know I'm weird. If I had 25 quarters, how much money would I have? Well, let's see, that'd be $6 and 25 cents. Yeah, that's it. Okay, now the next term, well, there's a negative over here. Let's bring that up to the front. Five cubed, that's 125 times four. That's 500, 500 Y cubed. Well, the next term be positive or negative? Positive. Positive, because the negative one is squared. And this is 25 times six. Well, if you had six quarters. You'd have $1.50. You'd have $1.50, I wish. Okay, so that would be 150 Y squared. The next term will be negative. Okay, are you sitting down? Good. What's five times four? Your bet is 20, 20 times Y. And then plus one. Well, let's see, does anybody see a shortcut that would explain what happens when you have a difference rather than a sum in this answer? Alternating signs. The signs alternate. So rather than putting in a negative one and a negative one and a negative one, what you could do is expand this using five Y and plus one. If you just remember to alternate the signs. You start off with a plus. And in this case, I happen to end up with a plus. But for other powers, I could end up with a negative on the end. Just go plus, minus, plus, minus, plus. And wherever the alternating signs lead you, so be it. So this is the expansion of five Y minus one. Now, you know, those of you at home and those people in the class may say, well, now Dennis, you know, we probably could have multiplied that out by hand faster than it took you to write all this down. But you know, I've been trying to explain it as I go. And there have been several of us contributing information here. I've been adding all the fancy jokes along the way. You can't forget that. So that's why it took me a little longer than usual to work this out. But I think with practice, Pascal's Triangle gives you a real advantage, especially if you have a power bigger than four. A fourth power isn't really that complicated to compute. But what if it had been a fourteenth power or a fourteenth power? You merely have to write Pascal's Triangle out to the appropriate row. And then you have to have a wide enough sheet of paper or enough lines to record all those terms on. But it only takes one line to write it out in factored form and the second line then to write it out in its simplified form. Okay, let's go back to the example screen and I'll try to remember to save Pascal's Triangle here. Now, see in the last example, what makes it more complicated is not so much the power. It's only a third power. But inside I've got a u cubed minus ten v squared. Let's try working that one out here on the green screen. We have u cubed minus ten v squared, and this is raised to the third power. Okay, well the first thing I would say is there are going to be how many terms in this answer? Four. Four terms, yeah, because this is a third power. So let's make room for four terms. Okay, now the next thing is there's a negative in here. So what can you tell me about the signs? They're going to alternate. So let's go back and make a negative here and make a negative over here. Okay, now I'm going to be taking these terms to be a and b. So I'll begin with my first term cubed. By the way, I'm using this row right here. So I'll have one times u cubed cubed. And then I'll have a three times u cubed squared times ten v squared. And then I'll have another three and a u cubed to the first power and a ten v squared squared. Let's move that negative over a little bit. And the last term will be ten v squared cubed. Okay, so I have my terms written out. That's my first stage. And then finally I go back and reduce these. What's u cubed cubed? Is it u to the 27th? U to the ninth. It's u to the ninth, thank you. Yeah, u to the 27th is a common wrong answer. Minus, now here I have a three and a ten. That makes a 30. Rather, u to the sixth and v squared. Plus, what will be the next coefficient? 30. Not 30. 300. Yes, because we have a ten squared times three. Plus 300 u cubed v to the fourth. v squared squared. And then minus 1,000 v to the sixth. So this is the expansion of this binomial. And I think that most people would find this approach quicker, especially for larger powers, that is using Pascal's triangle, than to do straight multiplication for these sorts of problems. Okay, well now we know how to use Pascal's triangle, but there's some more subtleties to it that we haven't investigated yet. Let's go to the next graphic and we'll introduce factorials. Now we'll be using factorials in the rest of this episode and in the next two episodes when we talk about permutations and combinations and probability. But right now I'm just introducing some terms. Now if you take the product of the first n positive integers, that's denoted by n factorial. And we put an exclamation mark after the n. So this does not mean n. No, it means n factorial. So n factorial means 1 times 2 times 3 times 4 up to n. Let me just use this space right here to take an example. Suppose I had 4 factorial. Well then that would mean 1 times 2 times 3 times 4. Which is how much? 24. Yeah, it looks like that's 6 times 4 is 24. Times 24. On the other hand, what if I wanted to change this to 5 factorial? What would I do differently? Times 5. Have to put a 5 in there, yeah. Okay, now how much is this? Well, let's see, here's a 12. And here's a 10. 12 and 10, the one's not going to affect the answer. That's 120. So we've gone from 24 to 120. If you were to go to 6 factorial, that would be 720. And if you go to 8 factorial, you're into the thousands. And so these factorials grow very rapidly. And many of you have a factorial key on your calculator. You may have, if you look at your calculator, you may have a key that looks like this. It may say x factorial with a little exclamation mark by it. And if you enter a 5 and then you push that key, immediately you should see 120. Now if this isn't written on a key, it may be written above a key on your calculator. Or it may be written under a menu, like if you have a TI-82 or a TI-89, you may see x factorial written under a menu that you could call up. And if you press that button, it will immediately multiply, in this case 1 through 5, and give you 120. Now you might say, well, why would my calculator have a factorial key on it? I've never used a factorial before. Well, maybe not so far, but in later courses, factorials are used sometimes, quite a bit. And in fact, in the rest of this course, we're going to be using factorials. So if you have this key available, you might keep it in mind. Okay, now, just to complete this graphic, there's one special case, and that's 0 factorial, or 0 with an exclamation mark. And this is defined to be 1, so we say that 0 factorial is equal to 1. Now, what's wrong with the definition right here? Why couldn't I just write 0 factorial using this definition? Because it doesn't start at 0. It doesn't start at 0. It starts at 1, and it goes up. So if you put a 0 in here, you can't start at 1 and go up to 0. So you see, this technically doesn't make any sense. But we define 0 factorial to be 1 so that it will fit with some of the some of the formulas and definitions that we see later. So we'll just take this to be 1. That's just the definition of it. Okay, now with this introduction to factorials, let's go to the next graphic and look at what are called binomial coefficients. For each pair of integers in an R where R is somewhere between 0 and N, so we're talking about integers. These would have to necessarily be non-negative integers because they're both at least 0 and R is less than or equal to N. Then we will be writing some expressions where I put parentheses and I stack two numbers in here, N and R. So if N is larger, I'll be putting the N on top. And the way we define this expression is we take the upper number factorial and I divide it by R factorial times the difference factorial. Now you might say, why in the world would you want to introduce something like this? Well, it's going to become quite useful here in the rest of this episode. It may not be something you've seen before, but it's something that you will be seeing now. Okay, if I use that definition, let's go to the green screen and calculate a few values that look like this. Suppose I had 4, 2. 4, 2. I'd like to evaluate this expression. This is sometimes referred to as a binomial coefficient, which we'll explain in a moment why it has that name. Now according to the definition, I'm going to put 4 factorial over 2 factorial times the difference factorial. 4 minus 2 factorial. That'll be 4 factorial over 2 factorial times 2 factorial again. Now to evaluate this, what I would do is to cancel off two of the factorials. You see in the numerator I have 1 times 2 times 3 times 4. In the denominator I have 1 times 2 and I have another 1 times 2. So I'm going to cancel off two portions of the factorials, this and this. And so what I'm left with is 3 times 4 over 2, which is 6. So this expression has a value of 6. And I'm using the formula, I probably should have written this again at the beginning, I'm using the formula n factorial over r factorial times n minus r factorial. Okay, I think we should do one more like this. What if we had, oh, let's see, what if we had 10, 7, 10, 7? Well, let's see, this would be 10 factorial over what? 7 factorial times 3 factorial times the difference factorial, 3 factorial. Okay, now on top I have how many factors? How many factors are there in 10 factorial? 10. 10 factors. Why don't I cancel 7 of those factors with 7 factorial? Because there is a 7 factorial within the 10 factorial. And if I cancel off 1 through 7, what would be left on top? 10 times, or 8 times 10. Yeah, or 10 times 9 times 8, I think Jeff was going to say, or 8 times 9 times 10. So I've canceled off the 7 through 1 and I've canceled off down here, 7 down to 1. And I'm left with 3 factorial on the bottom. So, you know, 3 factorial, if I were to write that out, would be 1 times 2 times 3. And the 2, I can cancel with the 8 and leave a 4. And the 3, I can cancel with the 9 and leave a 3. And so this answer is how much? 120. 120. Yeah. Now, you know, just above it here, let me just erase this upper portion and ask you a very similar question. Suppose I wanted to calculate this value, 10, but instead of 7, I'm going to put the difference in there, 3. 10 minus 7 is 3. What if I put in a 3 instead of a 7 there? So I'm replacing 7 with its, shall we say, with its complement. This will be 10 factorial over 3 factorial times what? 7 factorial. Times 7 factorial, because now 7 becomes the difference. Well, if you compare these two, these are exactly the same expressions. And therefore, this should also equal 120. So I think what we're demonstrating is a fact about factorials that I can summarize in this way. And that is that if I have n, r, if I have the binomial coefficient n, r, this is the same thing as n and put the complement of r on the bottom, n minus r. In other words, if I subtract 7 from n, rather r from n, I get n minus r. These numbers are the same thing. Let's write these out and see why. n, r is n factorial over r factorial times n minus r factorial. That was the definition. And the other expression, n, n minus r, is n factorial over n minus r factorial, that's the number that I have on bottom, times the difference. Now the difference is n minus n minus r factorial. So here I take my lower number n minus r factorial, and I take the difference. But you know if I reduce this, that's just going to be an r factorial. n factorial, n minus r factorial, and r factorial. Now if you compare this answer with that answer, those are the same. So we have this fact right here that says that n, r is the same as n, n minus r. We're going to see how this comes up in just a moment. In fact, let's look at it right now. Let's go to the next graphic on the binomial theorem. Okay, now this is a theorem that was first proved by Isaac Newton in a more general form. And it says that if you want to expand a plus b to the nth power, then you're going to get an a to the n, you're going to get an a to the n minus 1b, you're going to get an a to the n minus 2b squared. You notice here my powers on a going down, the powers on a go down by 1, the powers on b are going up. There aren't any b's, b to the first, b squared, all the way up to, finally, b to the n, and the a's are gone. Every term here has degree n. I have n variables there. I have n variables here, n variables here, all the way down. But now my coefficients do not come directly from Pascal's Triangle. These are the binomial coefficients that we just introduced. In my first term, I have zero b's, and so I call this n zero. In my next term, I have 1b, and I call this n1. In the next term, I have 2b's, and the binomial coefficient is n2. And so it goes. When I get to the n minus, when I get to the term that has n minus 1b's, it's n, n minus 1. And when there are nb's, I have the term nn. Okay. So, if I calculate each one of those binomial coefficients, that will tell me the number that came out of Pascal's Triangle. If we come to the green screen, let's just write out something like this using binomial coefficients. Suppose I had a plus b cubed. Now, we already know what the answer to that is. We've seen it come up today. But I'm going to write this as three zero, because n is three a cubed, plus three one a squared b. What would be the next coefficient? Three two. Three two, and what would be the variables? A b squared. A b squared. And finally, three three b cubed. Now, you know, this number and this number should be the same, because what I've done is I've replaced one with its complement two. That's the complement in three. So I've replaced one with the two. Those numbers should be alike. These numbers should be alike, because I've replaced zero with its complement in three. That would be a three. Okay, now, how much are each one of these numbers? I think I only need to calculate that one and that one, and then these two will be the same as the two that I've just calculated. So how much is three zero? Well, that's three factorial over zero factorial times the difference factorial, three minus zero factorial. That's three factorial over zero factorial times three factorial. Well, these two cancel off. And how much did we say zero factorial is defined to be? One. We call it a one. Yeah, just so that at moments like this, we can evaluate it and I get one over one or one. So the first coefficient is a one. Now, what's the coefficient here? What would you guess this should be when you cube A plus B? What should be the second coefficient? I think it should be a three. I think that's what Pascal's triangle told us. So this, if I calculate three one, there will be three factorial over one factorial times the difference factorial. That's two factorial. Now, to reduce this, let's cancel off the two factorial and what will be left on top? Three. There's a three. Yeah, we're canceling off the one and the two and the one and the two. There's a three left on top over one factorial, which is three over one, which is three. So what I can conclude from this is that A plus B to the third power should be one A cubed, so I'll write A cubed, plus three A squared B and then because of symmetry, this coefficient should be the same as that one because I have the complement as my lower number. So that'll be three A B squared and then this binomial coefficient should be the same as the first one because I've replaced zero with its complement. That'll be one B cubed and that's exactly what Pascal's Triangle gives me. We've seen this answer before. Now, you might say, well, Dennis, what is the point of this if we can use Pascal's Triangle to expand binomial? Why would I want to calculate the coefficients in this way? Well, you know, what if I weren't wanting to compute A plus B to the third power? What if I wanted to compute A plus B to the 30th power? Then I would have 31 terms and when I write down Pascal's Triangle, including the zero row, I'd have to write 31 rows all the way down. An alternative would be to calculate individual coefficients in this manner. We won't do A plus B to the 30th power, but this gives me a way of picking out individual coefficients and computing them directly without having to write out all of Pascal's Triangle. Okay. Let me just take another example like this and let's see how computing binomial coefficients can help me evaluate a problem. Suppose we had, let's say, 10C minus 2 to the fifth power. 10C minus 2 to the fifth power. Can anyone tell me how many terms this is going to have in this expansion? Six. It'll have six terms. It has one more term than the fifth power because I have to have a C to the fifth, a C to the fourth, third, squared, first power, and then the zero power. So there are actually six terms to write down. Now the first term will be 5,0 times 10C to the fifth power. And now the signs are going to alternate because of the negatives. I'll put a negative here. 5,1, 10C to the fourth power times 2. And then I'll put a plus. And then 5,2, 10C to the third power times 2 squared, minus. Okay, I'll just bring that minus down to the next row. Minus 5,3 times 10C squared times 2 cubed. You notice the degree is always going to be a total of five, five factors there. Plus 5,4, 10C to the first power times 2 to the fourth. Minus 5,5 times 2 to the fifth because my 10Cs now drop out. Okay, now let's see. These two coefficients will be the same. These two coefficients will be the same. And these two coefficients will be the same. I'm using that symmetry property. These are alike, these are alike. And now I'll do a little crisscross. These are alike. So I really only need to calculate three coefficients, not five. Let's calculate 5,0. I bet you know what that's going to be. 5,0, that's 5 factorial over 0 factorial times 5 factorial. This being the difference right here and I get a 1. Yeah, so that lead term should have a 1 on it. And then 5,1 is 5 factorial over 1 factorial times the difference factorial. And if I cancel off 4 factorial, I have a 5 left on top. 5 over 1 is 5. And then if I go to the coefficient here, 5,2, I'll have 5 factorial over 2 factorial times the difference 3 factorial. What would you cancel off here to make this simpler to compute? What would be better to cancel the 2 factorial or the 3 factorial? The 3 factorial. Yeah, let's cancel as many as we can. Let's cancel 3 factorial and that leaves me with 4 times 5. But I still have the 2 factorial, 1 times 2. That's 20 over 2 is 10. So that tells me the first coefficient here is a 1. The second coefficient is a 5. The next coefficient is a 10. But now what's this coefficient? 10. That's a 10 because, see, I'm using the complement of the 2 that came just before it. That's a 10. This is a 5 because it agrees with that one and this one is a 1. Okay, I have all my coefficients. So, excuse me. So now I'm going to expand this and write down my final answer. Let's see, 10 c to the fifth power. What's 10 to the fifth? 100,000. 100,000. Yeah, it's a 1 with 5 zeros. 100,000 c to the fifth. Minus. Now next I have 10 to the fourth. That's going to be, what's 10 to the fourth? 10,000. 10,000. But I have another 10, a 5 times 2. So that makes 100,000 again c to the fourth. Plus. And now I have 10 to the third. Let's see, 10 to the third, that's 1,000. Times 10 is 10,000. Times 4 is 40,000. That's 40,000 c cubed. Minus. Okay, now the next term. This is going to be 100, because there's a 10 squared there. Times 10 to the thousand, times 8 is 8,000 c squared. Plus. Let's go down to the next line. 10 times 5. I'll tell you what, let's take one of the 2's in that 5. One of these 2's in that 5's makes a 10. There's another 10, that's 100. And I have 3 2's left over. That's 100 times 8, 800 c. Minus. Now all my 10's are gone. 2 to the fifth is 32. So this is 10 c minus 2 to the fifth power. It's this 6 term polynomial. Now you might say, well you know Pascal's triangle might have made this a little bit quicker. But I do have an alternate way now of picking out individual terms and calculating them. And so this gives me now a second way to compute these binomials. Okay, let's go to the last graphic. Here I ask you a few questions about the expansion of a binomial. And I think we've been answering questions sort of like this, but here we see it in a concrete form. How many terms are there in the expansion of a plus b to the 12th power? 13. There would be 13 terms. Yeah, 13 terms. Because it's a 12th power and we get an extra power to include an extra term because there's a zero power involved. What's going to be the very first term of that expansion? Jeff, what would you say? A to the 12th. A to the 12th. Yeah, it would be 1 times A to the 12th. And what's going to be the 13th term? B to the 12th. B to the 12th. Yes, that's going to be done at the far right end. B to the 12th. Now what's the fifth term going to be? Well, let's see now. The fifth term is going to be, we started off with A to the 12th and we're going to reduce it four times so we'll have A to the what? A to the 8th, I think. And the B's, which started on zero, are going to come up to B to the 4th. So it would be A to the 8th, B to the 4th. Now what we need is a coefficient. Let's come to the green screen and try calculating the fifth term. So let's write down that question. We're wondering what is the fifth term of A plus B to the 12th power? Well, let's see. We have a term and we have a term and we have a term and we have a term. Here's the fifth term. That's the one we want. Now what we found out was the very first term is A to the 12th and we found out that the very last term down here is B to the 12th. But the question is what is the fifth term right there? Well, that term is going to have to have an A to the 8th and a B to the 4th. Now the way I figured that is we started on 12 and we got 11, 10, 9, 8. 8 to the 8th. And the B's are going up. B to the 0, 1st, 2nd, 3rd, 4th. B to the 4th. And the coefficient will have what two numbers in the binomial coefficient? Twelve and... Yeah, what number are we going to put there? Four. Twelve and four. Because you see the very first coefficient is 12, 0. So here's 12, 0, 12, 1, 12, 2, 12, 3, 12, 4. So you can't go by the fifth term and put a 5 here. You go by... Well, what I go by is the power on the B. So B, 12, 4. So how much is 12, 4? Well, let's just write it out. That's 12 factorial over 4 factorial times 8 factorial A to the 8th, B to the 4th. Now, if you were going to cancel here, which of the denominator factors would you cancel? Eight factorial. Eight factorial. Get rid of as many as you can. So that's going to leave us with 9, 10, 11, and 12 on top. Because we canceled 1 through 8. On the bottom I have 1, 2, 3, and 4. Well, let's see. 3 and 4 is 12. So that gets rid of that guy. Whether than dividing by 2, let's go ahead and multiply this out and divide by 2 afterwards. I'm thinking this is 99 times 10 is 990 over... Oh, you know what? I left off my variables. Let me just put my variables on the end of that. A to the 8th, B to the 4th. Now, when I multiply that out, I get 990 over 2, A to the 8th, B to the 4th. And now let's divide by 2, so I don't have to do any further multiplication. 4, 95, A to the 8th, B to the 4th. That is the 5th term. That's what goes right here. That's going to be 4, 95, A to the 8th, B to the 4th. I have to squeeze that in that position. Okay, one more example. And I'll do this one on the green screen. I'm wondering how much is 11 raised to the 6th power? 11 raised to the 6th power. Now, I'll tell you what. Let me make that a little bit smaller. That may be a little bit too aggressive there. Let's go to the 4th power, and then we'll do the 6th power after this. Now, in this case, I'm going to go back and use Pascal's Triangle one more time to illustrate, just to remind people how this goes. And I'm going to raise this number to the 4th power using Pascal's Triangle. The way I'll do it is I'll think of this as being 10 plus 1 to the 4th power. So this is my A plus B. So when I expand this, I get 10 to the 4th plus 4 times 10 to the 3rd times 1 plus 6, oops, better put a multiplication dot in there, plus 6 times 10 squared times 1 squared. I won't bother writing that one out. 1 squared is just 1. Plus 4 times 10 plus 1. Now, you notice this is an abbreviated form. I didn't show my B factors because those are all 1s. And so this is going to give me 10,000 plus 4,000 plus 600 plus 40 plus 1. Now, if you add that all up, that's going to be 10,000, 4,641. Where have you seen these numbers before? In Pascal's Triangle. That is Pascal's Triangle. You see that row of the triangle represents 11 raised to that power. What would you guess would be 11 cubed? 1,331. 1,331. Here's 11 squared. Look, here's 11. And here's 11 to the 0 power. Well, it was nice seeing you today, and I'll see you next time for permutations and combinations in episode 37.