Algebra / choosing, multiplying, counting

Where do the coefficients come from?

What happens when the same bracket is multiplied by itself many times, as in $(a + b)^5$ ? Writing out every product soon takes a long time, yet expansions like this keep appearing. For example, finding the slope of a curve such as $y = x^3$ needs $(x + h)^3$ , the same kind of expression that the derivative page uses for $x^2$ . Counting problems lead to the same structure, because each bracket offers a choice between two terms. To expand two brackets, we multiply every term in one bracket by every term in the other (the factorisation page uses this). Here we look for a quicker way to handle many brackets. That slope example is only one place where these expansions appear; if it is new to you, skip it for now. What follows starts from bracket multiplication and counting.

Let us begin with $(a + b)^2$ , which means $(a + b)(a + b)$ . Why does this give $2ab$ rather than just $ab$ ? We will write out the four products first, then add one more bracket. Keeping track of these choices will help us find the pattern for larger powers.

aone termbthe other termLetters are numbers, not objects being drawn at random.

01Two brackets. Four selections.

The distributive law says $x(a+b)=xa+xb$ : multiply by each term and add. Apply it to both brackets. Take a or b from the first bracket, then a or b from the second. Play the animation, pause it, or collect one product at a time.

Choose → multiply → collect0 / 4 selections collected

The exponent 2 means 2 copies of (a+b), multiplied together. To expand, take one letter from every bracket and multiply those letters.

Animation progress0%

Collect every selection to assemble the expansion.

The selections ab and ba come from different bracket positions. But ordinary numbers satisfy ab = ba, so their products are equal. Adding them gives ab + ab = 2ab. The 2 in a² counts factors of a; the 2 in 2ab counts equal products being added.

02Three brackets. Count the positions.

Now each selection uses three letters, so its exponents always add to 3. To obtain $a^2b$ , two brackets must supply a and the remaining bracket must supply $b$ . Watch all eight selections, then compare the three ways to obtain $a^2b$ .

Choose → multiply → collect0 / 8 selections collected

The exponent 3 means 3 copies of (a+b), multiplied together. To expand, take one letter from every bracket and multiply those letters.

Animation progress0%

Collect every selection to assemble the expansion.

Why does $a^2b$ appear three times?

The brackets occupy three distinct positions. Choose two positions for a, and the remaining position must supply b. Equivalently, choose just that one b position.

Choose position 1, 2

Choose position 1, 3

Choose position 2, 3

\binom{3}{2}=\binom{3}{1}=3

These are two ways to count the same three selections, not six different selections. A combination counts which positions you choose; it does not count the order in which you click them.

{}^nC_r=\binom{n}{r}

Read this as “n choose r”. $n$ is the total number of available positions; $r$ is how many positions to choose. Here $n = 3$ and $r = 2$ (for $a$ ), or $r = 1$ (for $b$ ).

03 $n$ brackets. Choose $k$ of them for $a$ .

Let $k$ count the brackets that supply a. Every other bracket supplies $b$ , giving $n-k$ copies of $b$ . The product is $a^kb^{n-k}$ . Its coefficient is “n choose k”, because any selection of those $k$ positions produces that same product. $k$ now plays the role of $r$ in $nCr$ .

Every coefficient is a countn = 5, k = 2

n · total brackets5

k · brackets choosing a2

a \cdot a \cdot b \cdot b \cdot b=a^{2}b^{3}

2 copies of a and 3 copies of b. Click any bracket above to change its choice.

$\binom{5}{2}$ = 10 different selections give this same product.

COUNT

Choose 2 positions from 5 for a. The remaining 3 positions must supply b. There are 10 such choices, so the coefficient of a²b³ is 10.

(a+b)⁵ =+++++

The terms run from zero a’s to n a’s. Click a term to inspect every selection behind its coefficient. Reordering the terms does not change their sum.

(a+b)^n=\binom n0 b^n+\binom n1 ab^{n-1}+\cdots+\binom nk a^k b^{n-k}+\cdots+\binom nn a^n

Start with $k=0$ : take $b$ from every bracket, giving $b^n$ once. Increase $k$ one at a time, until $k=n$ : take a from every bracket, giving $a^n$ once. This lists every selection exactly once, grouped by its product.

\boxed{(a+b)^n=\sum_{k=0}^{n}\binom nk a^k b^{n-k}}

$\Sigma$ means “add these terms”. The bottom $k=0$ tells you where to start; the top $n$ tells you where to stop. The powers tell you what was chosen; the coefficient tells you how many ways it could be chosen. This is the binomial theorem.

Here $n$ is a nonnegative integer. A factor with exponent zero is omitted; for $n=0$ the expansion is 1. The interactive display reaches $n=8$ , but the counting argument works for every nonnegative integer $n$ .

Where do the coefficients come from?

01Two brackets. Four selections.

02Three brackets. Count the positions.

Why does $a^2b$ appear three times?

03 $n$ brackets. Choose $k$ of them for $a$ .

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01Two brackets. Four selections.

02Three brackets. Count the positions.

Why does a2ba^2ba2b appear three times?

03nnn brackets. Choose kkk of them for aaa.

What would you like to do next?

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Why does $a^2b$ appear three times?

03 $n$ brackets. Choose $k$ of them for $a$ .