The Binomial and Multinomial Theorems

You have seen binomial expansion as an algebra drill. Combinatorics reveals what those coefficients are: they are counts. Expanding (x+y)^n is secretly a counting problem, and once you see that, the whole theorem — and its many-variable big brother — becomes obvious rather than memorised.

Write (x+y)^n = (x+y)(x+y)\cdots(x+y), n factors. To expand, you go through the brackets and from each one pick either an x or a y, then multiply your picks. Every term of the product is one such choice-sequence. A term with exactly k of the y's (and hence n-k of the x's) equals x^{n-k}y^k — and the number of ways to get it is the number of ways to choose which k brackets donated the y: that is \binom{n}{k}. Collect like terms and the coefficient of x^{n-k}y^k is exactly \binom{n}{k}.

For any n \ge 0,

(x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k = \binom{n}{0}x^n + \binom{n}{1}x^{n-1}y + \cdots + \binom{n}{n}y^n.

The coefficient of x^{n-k}y^k is the count \binom{n}{k}.
Setting x = y = 1 gives \sum_k \binom{n}{k} = 2^n; setting x=1, y=-1 gives \sum_k (-1)^k\binom{n}{k} = 0.

Pascal's triangle: the coefficients build themselves

Stack the coefficient rows and you get Pascal's triangle: row n is \binom{n}{0}, \binom{n}{1}, \dots, \binom{n}{n}. Every inner entry is the sum of the two directly above it, which is precisely the identity \binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k} — fix one bracket and split on whether it gave a y or not. Drag the slider to add rows and watch the sums propagate.

So you never need to memorise coefficients: (x+y)^4 reads straight off row 4 as 1, 4, 6, 4, 1, giving x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4. The triangle is a machine that grows binomial coefficients by addition alone.

Worked examples

A single term, without expanding everything

Find the coefficient of x^5 in (2 + x)^8. The general term is \binom{8}{k} 2^{8-k} x^{k}; take k = 5:

\binom{8}{5} 2^{3} = 56 \cdot 8 = 448.

The point of the theorem is exactly this — you can pluck one coefficient out of a huge expansion without writing the other terms.

A quick estimate

(1.01)^{10} = (1 + 0.01)^{10} \approx 1 + 10(0.01) + \binom{10}{2}(0.01)^2 = 1 + 0.1 + 0.0045 = 1.1045, already accurate to four decimals. The early binomial terms are the workhorse behind linear and quadratic approximations everywhere in applied maths.

Many variables: the Multinomial Theorem

What if there are three or more terms in the bracket? Expand (x_1 + x_2 + \cdots + x_m)^n by the same story: from each of the n brackets pick one of the m variables. A term x_1^{k_1} x_2^{k_2} \cdots x_m^{k_m} (with k_1 + k_2 + \cdots + k_m = n) is obtained by choosing which brackets give each variable — the number of ways to arrange a word made of k_1 copies of "1", k_2 copies of "2", and so on. That count is the multinomial coefficient:

\binom{n}{k_1, k_2, \dots, k_m} = \frac{n!}{k_1!\,k_2!\cdots k_m!}.

(x_1 + \cdots + x_m)^n = \sum_{k_1 + \cdots + k_m = n} \frac{n!}{k_1!\cdots k_m!}\, x_1^{k_1}\cdots x_m^{k_m}.

The m = 2 case is the binomial theorem: \frac{n!}{k!\,(n-k)!} = \binom{n}{k}.
Setting every x_i = 1 gives \sum \binom{n}{k_1,\dots,k_m} = m^n — the product-rule count of all length-n words over an m-letter alphabet.

Example: the coefficient of x^2 y^2 z in (x + y + z)^5 is \frac{5!}{2!\,2!\,1!} = \frac{120}{4} = 30. And the number of distinct arrangements of MISSISSIPPI (11 letters: 1 M, 4 I, 4 S, 2 P) is the multinomial coefficient \frac{11!}{1!\,4!\,4!\,2!} = 34{,}650 — the same object, wearing an anagram costume.

Each term is a solution of k_1 + k_2 + k_3 = n in non-negative integers. Counting those is a "stars and bars" problem: lay down n stars and 2 bars to split them into three groups, so the number of distinct terms is \binom{n+2}{2} = \frac{(n+1)(n+2)}{2}. For a general m-variable power it is \binom{n+m-1}{m-1}. So (x+y+z)^4 has \binom{6}{2} = 15 distinct monomials — a lot more than the binomial's n+1.

The coefficient is not just \binom{n}{k} when the terms carry constants. In (2+x)^8 the coefficient of x^5 is \binom{8}{5}\,2^3, not \binom{8}{5} — the theorem is about (x+y)^n and here y = x, x = 2. Always match the term to \binom{n}{k}x^{n-k}y^k including the powers of the constants.
Powers must sum to n. A multinomial term x_1^{k_1}\cdots x_m^{k_m} only exists when k_1 + \cdots + k_m = n; ask for a coefficient whose exponents don't add to n and the answer is simply 0.
Signs. For (x - y)^n write it as (x + (-y))^n: the term is \binom{n}{k}x^{n-k}(-y)^k, so odd k terms come out negative. Dropping the sign is the single most common expansion error.