The Binomial Theorem

May 2026 · 16 min read

The binomial coefficients we built in Part 4 are not just counters of subsets. They are also the coefficients you get when you expand $(x + y)^n$. The connection between the algebra and the combinatorics is the binomial theorem — and once you see why the bridge exists, almost every "messy expansion" problem becomes routine.

Statement

Equivalently, you can index by the power of $x$ and write

$\displaystyle (x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{k} y^{n-k},$

using the symmetry $\binom{n}{k} = \binom{n}{n-k}$.

Combinatorial Proof

Write $(x + y)^n$ explicitly as the product of $n$ identical factors:

$(x + y)(x + y)(x + y) \cdots (x + y) \quad\text{($n$ factors).}$

To expand the product into a sum of monomials, the distributive law tells you to pick either $x$ or $y$ from each factor and multiply your choices. A typical term in the expansion looks like $y \cdot x \cdot x \cdot y \cdot x \cdot \dots$, which after rearrangement is some power of $x$ times some power of $y$.

Suppose we end up choosing $y$ from exactly $k$ of the $n$ factors and $x$ from the remaining $n - k$. The product is then $x^{n-k} y^k$. How many ways can we make those choices? We are picking $k$ of the $n$ factors to take $y$ from — that is exactly the number of $k$-subsets of an $n$-set, namely $\binom{n}{k}$.

So the coefficient of $x^{n-k} y^k$ in the expansion is $\binom{n}{k}$. Summing over all valid $k$ gives the binomial theorem. $\Box$

Worked: $(x + y)^4$

Read the row $n = 4$ of Pascal's triangle: $1, 4, 6, 4, 1$. So

$(x + y)^4 = x^4 + 4x^3 y + 6 x^2 y^2 + 4 x y^3 + y^4.$

You can verify the coefficient of $x^2 y^2$ directly: we need to pick $y$ from 2 of the 4 factors. There are $\binom{4}{2} = 6$ ways to do that. Pleasingly, that is the same 6 you would compute by brute multiplication.

Useful Substitutions

Setting $x = y = 1$

$\displaystyle 2^n = (1 + 1)^n = \sum_{k=0}^{n} \binom{n}{k} \cdot 1^{n-k} \cdot 1^{k} = \sum_{k=0}^{n} \binom{n}{k}.$

This is the row sum we mentioned in Part 4. The binomial theorem turns it from a curiosity into a one-line proof.

Setting $x = 1, y = -1$

$\displaystyle 0 = (1 - 1)^n = \sum_{k=0}^{n} \binom{n}{k} (-1)^k$ for $n \ge 1$.

Splitting into even-$k$ and odd-$k$ sums shows that there are equally many even-sized and odd-sized subsets of an $n$-set when $n \ge 1$.

Differentiating with respect to $x$

Treat $y$ as a constant. Differentiate both sides of the identity $(1 + x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k$ with respect to $x$:

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

Set $x = 1$ to get

$\displaystyle \sum_{k=0}^{n} k \binom{n}{k} = n \cdot 2^{n-1}.$

This is also the answer to "pick a committee from $n$ people and designate a chair" counted in two ways: $\binom{n}{k}$ for the committee, $k$ for the chair, summed over all $k$ — versus $n$ for the chair (any of the $n$ people), then $2^{n-1}$ for the rest.

Integrating

Integrate $(1 + x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k$ from $0$ to $1$:

$\displaystyle \dfrac{2^{n+1} - 1}{n + 1} = \sum_{k=0}^{n} \dfrac{\binom{n}{k}}{k + 1}.$

Both calculus and combinatorics deliver the same identities — pick whichever proof you find faster.

Beyond Two Variables: A Preview

What about $(x + y + z)^n$? Each factor in the product offers three choices instead of two, so the expansion is

$\displaystyle (x + y + z)^n = \sum_{a + b + c = n} \binom{n}{a, b, c} x^a y^b z^c$

where $\binom{n}{a, b, c} = \dfrac{n!}{a!\,b!\,c!}$ is a multinomial coefficient. We tackle these in Part 6.

A Ross-style Application

How many subsets of $\{1, 2, \dots, n\}$ have an even number of elements?

From the alternating-sum identity above, $\sum_{k} (-1)^k \binom{n}{k} = 0$. Splitting odd and even $k$:

$\sum_{k\text{ even}} \binom{n}{k} = \sum_{k\text{ odd}} \binom{n}{k}.$

And the total is $\sum_k \binom{n}{k} = 2^n$. So each side equals $2^{n-1}$. There are exactly $2^{n-1}$ even-sized subsets when $n \ge 1$.

Heads-up: General Binomial Theorem

For non-integer or negative $\alpha$, the same theorem works as a power series in $|x| < 1$:

$(1 + x)^\alpha = \sum_{k=0}^{\infty} \binom{\alpha}{k} x^k,$

where the generalised binomial coefficient is

$\binom{\alpha}{k} = \dfrac{\alpha (\alpha - 1)(\alpha - 2) \cdots (\alpha - k + 1)}{k!}.$

This is how Newton extended the theorem to fractional and negative powers. We will not need it for elementary probability, but it is good to know it exists — it powers Taylor expansions of $\sqrt{1 + x}$, $1/(1 - x)$, and a great deal of generating-function algebra.

Common Mistakes

• Swapping $x^{n-k}$ and $y^k$. Always check which variable's exponent matches the binomial coefficient subscript. Both conventions appear in the wild — pick one and stick with it.
• Treating $(x - y)^n$ as $(x + y)^n$. When $y$ is replaced by $-y$, every odd-power term flips sign. The expansion is $\sum (-1)^k \binom{n}{k} x^{n-k} y^k$.
• Dropping the $\binom{n}{0}$ or $\binom{n}{n}$ term. They are both 1 and easy to forget when summing manually.
• Assuming the formula needs $x, y$ to be numbers. The proof uses only the distributive law and commutativity of multiplication — it works for symbolic variables, matrices that commute, polynomials, and many other settings.

Summary

• $(x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k$.
• Coefficient of $x^{n-k} y^k$ is the number of ways to choose which $k$ factors contribute $y$.
• Substitutions like $x = y = 1$ or $x = 1, y = -1$ instantly produce row-sum and alternating-sum identities.
• Differentiation and integration of the identity $(1 + x)^n = \sum \binom{n}{k} x^k$ give weighted sum identities.
• Generalises to multinomials (Part 6) and to non-integer exponents via Newton's series.

Practice Problems

• Ross 1.24. Expand $(3x^2 + y)^5$.
• Ross 1.25. The game of bridge is played by 4 players, each of whom is dealt 13 cards. How many bridge deals are possible? multinomial coefficient preview
• Practice — sum identity. Prove $\sum_{k=0}^n k^2 \binom{n}{k} = n(n+1) 2^{n-2}$ for $n \ge 1$. differentiate twice
• Practice — even sum. Show that $\sum_{k=0}^n \binom{n}{k} 3^k = 4^n$ in two ways — algebraically and combinatorially.
• Practice — coefficient extraction. Find the coefficient of $x^{10}$ in $(1 + x + x^2)^{20}$. generating-function flavour; can be done with multinomials
• Practice — alternating sum with weight. Prove $\sum_{k=0}^n (-1)^k k \binom{n}{k} = 0$ for $n \ge 2$. differentiate the alternating identity
• LeetCode 50 — Pow(x, n). Implement fast exponentiation — useful background for evaluating $(1 + x)^n$ at integer points modulo a prime.