The Multinomial Theorem

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle to start. Take a plus b, to the power ten. It has eleven terms, right? Now add just one more letter: a plus b plus c, to the power ten. How many terms now? Twelve? Keep your guess in mind. Today: the multinomial theorem, the binomial theorem for more than two letters. It comes up often in JEE, so let's begin.

0:26One letter from each bracket

So, the binomial theorem had just two letters. Now take k letters: x one, x two, and so on up to x k, each with a small number written below the x. The three dots mean, and so on. Raise it all to the power n, where n is a positive integer. That is n brackets multiplied together, with k letters in each. Okay? Now think. When you multiply out these n brackets, how do you make one single term? Yes, exactly! Pick one letter from each bracket, and multiply your picks. If you pick x one in n one brackets, x two in n two brackets, and so on, the term is x one to the power n one, times x two to the power n two, up to x k to the power n k. One pick per bracket, so n one plus n two, up to n k, adds up to n. With me? But the same term comes from many different sets of picks. How many? As many as the ways to arrange n one copies of x one, n two copies of x two, and so on, across the n brackets. That is n factorial, over n one factorial, n two factorial, up to n k factorial. The exclamation mark means factorial. Clear?

1:32The multinomial theorem

So, put it together: every term, times the number of ways it arises, all added up. In one line: x one plus x two, up to x k, to the power n, equals the sum of these coefficients, each times its term. The big Greek letter sigma means, add them all up. This is the multinomial theorem. Okay? The sum runs over every choice of the powers n one, n two, up to n k, that add up to n. And a power may be zero: that letter was simply never picked. Right?

2:03Counting terms, and the sum of coefficients

Now, back to our puzzle: how many terms? One for each choice of the powers. When the x's are independent letters, that count is n plus k minus one, C, k minus one: n plus k minus one at the top left of the capital C, and k minus one below. Okay? For three letters, k equals three: n plus two C two, which is n plus one, times n plus two, over two. Clear? So, a plus b plus c, to the power ten. How many terms? Twelve? No! Here n is ten, so it is twelve C two: eleven times twelve, over two, where the dot means times. Sixty six terms, not twelve! Puzzle solved! And the sum of all the coefficients? Put every x i equal to one, where x with a small i below means any one of the letters. Each bracket becomes k, so the sum is k to the power n. With me?

2:55Coefficient of xᵐ in (a+bx+cx²)ⁿ

Now a case that comes up often: a, plus b x, plus c x squared, to the power n, all powers of the same x. We want the coefficient of x to the power m. Pick a in p brackets, b x in q brackets, and c x squared in s brackets. The term is n factorial over p factorial, q factorial, s factorial, times a to the power p, b x to the power q, and c x squared to the power s, each in round brackets. Right? So tell me, what power of x does this term carry? q from the b x part, and two s from the c x squared part: q plus two s, correct! So two conditions: p plus q plus s equals n, and q plus two s equals m. Clear? Then add n factorial over p factorial, q factorial, s factorial, times a to the power p, b to the power q, c to the power s, over every triple p, q, s, in round brackets, that meets both conditions. Got it?

3:52Trap: powers of the same variable

Now, a trap. One plus x plus x squared, to the power n. Three terms in the bracket, so n plus one, times n plus two, over two terms, right? No, no, careful! All three are powers of the same x. So different triples p, q, s can give the same power of x, and those terms combine into one. Okay? The powers run from x to the power zero, up to x to the power two n. So there are only two n plus one distinct terms. The earlier count works only for independent letters. Don't fall for it! Got it?

4:25Example 1: coefficient of x^2 y^3 z^4 in (2x - y + 3z)^9

Okay, time for an example. Find the coefficient of x squared, y cubed, z to the power four, in two x, minus y, plus three z, to the power nine. Pause the video and try it. Done? Okay, let's check. The powers two, three and four add up to nine, the index. So the multinomial coefficient is nine factorial, over two factorial, three factorial, four factorial: one thousand two hundred and sixty. Okay? So the answer is one thousand two hundred and sixty, right? No, no, wait! Here the letters carry numbers with them. We pick two x twice, minus y three times, and three z four times: two x, in brackets, squared; minus y, in brackets, cubed; and three z, in brackets, to the power four. With me? So multiply in two squared, which is four; minus one, cubed, which is minus one; and three to the power four, which is eighty one. Clear? One thousand two hundred and sixty, times four, times minus one, times eighty one: minus four hundred and eight thousand, two hundred and forty. The odd power of minus y made it negative. Very good!

5:36Example 2: coefficient of x^5 in (1 + x + x^2)^4

Now a harder one. Find the coefficient of x to the power five, in one plus x plus x squared, to the power four. Remember the trap we just saw. Pause the video and try it. Done? Okay, let's check. Here a, b and c are all one. We need p plus q plus s equals four, and q plus two s equals five. Okay? Try s equals one: then q is three, and p is zero. That triple, in round brackets, gives four factorial, over three factorial, times one factorial: four. Clear? So the coefficient of x to the power five is four, right? Oh no, careful! That is exactly the trap: another triple gives the same power. With s equal to two, q is one and p is one: four factorial, over one factorial, one factorial, two factorial, which is twelve. With me? And s equal to zero gives q equal to five, too many for four brackets; s equal to three or more makes q negative. Both impossible. So the total is four plus twelve: sixteen. Perfect!

6:43Quick revision: coefficient, number of terms, sum k^n

So, quick revision. The coefficient is n factorial, over the factorials of the powers, times the numbers the letters carry. Independent letters: n plus k minus one, C, k minus one terms, and the coefficients add up to k to the power n. Powers of one x: solve p plus q plus s equals n, and q plus two s equals m, and add every triple. Now go and try the practice problems on this page. All the best!