The Binomial Theorem

The Binomial Theorem is the seed of an entire branch of enumerative combinatorics. Setting $x=y=1$ gives $2^n = \sum_k \binom{n}{k}$, the fact that a set of size $n$ has $2^n$ subsets, unifying two seemingly different counting facts. Setting $x=1,y=-1$ produces the alternating-sum identity used constantly in inclusion–exclusion arguments. These specializations, and dozens more (Vandermonde's identity, the hockey-stick identity, generating-function derivatives), form the subject of Binomial Identities. In probability and statistics, the binomial distribution's probability mass function $\binom{n}{k}p^k(1-p)^{n-k}$ is literally a term of the expansion of $(p+(1-p))^n = 1$, and the theorem guarantees these probabilities sum to $1$. In algebra and calculus, the theorem underlies finite-difference calculus and the expansion of derivatives of products (Leibniz's rule for $\frac{d^n}{dx^n}(fg)$ has the identical coefficient pattern). In computer science, binomial coefficients count subsets in algorithm analysis (e.g., the number of size-$k$ combinations tried by brute-force search) and appear in the analysis of binomial heaps and random binary structures. The theorem is also the first and simplest instance of the theory of Ordinary Generating Functions: the polynomial $(1+y)^n$ is the generating function whose coefficient sequence is exactly the row of Pascal's Triangle $\binom{n}{0},\binom{n}{1},\dots,\binom{n}{n}$, and this viewpoint scales up to encode far more complex counting sequences (partitions, compositions, recurrences) as coefficients of more elaborate power series. In cryptography and coding theory, binomial coefficients count error patterns of fixed weight, central to the analysis of Hamming and Reed–Solomon codes' error-correcting capacity.

Ask the tutor: How does the Binomial Theorem generalize when you replace $(x+y)^n$ with $(x_1+x_2+\cdots+x_m)^n$?

Think of $(x+y)^n$ not as a formula to expand, but as $n$ physical boxes lined up in a row, box $1$ through box $n$, each containing a copy of $x+y$. To expand the product, you must reach into every box and pull out either its $x$ or its $y$, then multiply your $n$ choices together. Doing this for every possible combination of choices and adding up all the resulting products is exactly what "expanding" means — there is no other operation happening. Every one of the $2^n$ ways of making these choices produces a monomial $x^{n-k}y^k$ where $k$ is the number of boxes from which you drew a $y$. Many different choice-sequences produce the same monomial: any sequence with $y$ drawn from exactly the same $k$ boxes gives $x^{n-k}y^k$, and swapping which boxes contributed the $y$'s (among the same total count $k$) does not change the monomial at all, since multiplication is commutative. So the coefficient of $x^{n-k}y^k$ in the fully expanded, un-simplified sum is the number of ways to choose which $k$ of the $n$ boxes contribute a $y$ — precisely $\binom{n}{k}$.

The picture makes the coefficient tangible: there are $\binom{3}{1}=3$ rows with exactly one $y$, giving coefficient $3$ on $x^2y$, and $\binom{3}{2}=3$ rows with exactly two $y$'s, giving coefficient $3$ on $xy^2$, matching $(x+y)^3 = x^3+3x^2y+3xy^2+y^3$.

Arranging the coefficients $\binom{n}{k}$ by row $n$ and position $k$ produces the classical triangular array, where each entry is the sum of the two entries diagonally above it — this is the content of Binomial Identities such as Pascal's Rule, which we prove below as a lemma en route to (and independently of) the main theorem.

Example 1 — expanding $(2a - 3b)^4$. We apply the theorem with $x = 2a$, $y=-3b$, $n=4$. The coefficients for $n=4$ are $\binom{4}{0},\dots,\binom{4}{4} = 1,4,6,4,1$ (row 4 of Pascal's Triangle). Then

Compute each term explicitly:

• $k=0$: $\binom{4}{0}(2a)^4(-3b)^0 = 1\cdot 16a^4 \cdot 1 = 16a^4$.
• $k=1$: $\binom{4}{1}(2a)^3(-3b)^1 = 4\cdot 8a^3 \cdot(-3b) = -96a^3b$.
• $k=2$: $\binom{4}{2}(2a)^2(-3b)^2 = 6\cdot 4a^2\cdot 9b^2 = 216a^2b^2$.
• $k=3$: $\binom{4}{3}(2a)^1(-3b)^3 = 4\cdot 2a\cdot(-27b^3) = -216ab^3$.
• $k=4$: $\binom{4}{4}(2a)^0(-3b)^4 = 1\cdot 1\cdot 81b^4 = 81b^4$.

Summing,

As a sanity check, substitute $a=1,b=0$: the right side gives $16$, matching $(2\cdot1-3\cdot0)^4 = 2^4=16$. Substitute $a=0,b=1$: right side gives $81$, matching $(-3)^4=81$. Substitute $a=1,b=1$: right side gives $16-96+216-216+81 = 1$, matching $(2-3)^4=(-1)^4=1$ exactly — a strong consistency check since it exercises every term with cancellation.

Example 2 — extracting one coefficient from $(3 + x^2)^7$ without full expansion. Suppose we only want the coefficient of $x^{10}$ in $(3+x^2)^7$. Apply the theorem with the roles $x_{\text{thm}} = 3$, $y_{\text{thm}} = x^2$, $n=7$:

We need $2k = 10$, i.e. $k=5$. The coefficient is

Double-check $\binom{7}{5} = \binom{7}{2} = \frac{7\cdot 6}{2} = 21$ by the symmetry identity $\binom{n}{k}=\binom{n}{n-k}$ (itself an immediate consequence of the bijection "complement a $k$-subset to get an $(n-k)$-subset," which will be revisited under Binomial Identities). So the coefficient of $x^{10}$ in $(3+x^2)^7$ is exactly $189$, and this required no other term of the expansion — a genuine payoff of having the closed-form term $\binom{n}{k}x^{n-k}y^k$ rather than needing the full product.

The binomial coefficients and the expansion of $(x+y)^n$ for integer $n$ were known, in essentially complete generality, well before any single "discoverer": Indian mathematicians (Pingala, and later Halayudha's commentary describing the triangular array), Chinese mathematician Jia Xian and later Yang Hui, and Persian mathematician al-Karaji all worked with these coefficients and their triangular recursive structure centuries before European mathematics did. In Europe the array bears the name of Blaise Pascal, who in 1654 wrote a systematic treatise organizing the triangle's properties, its recursive rule, and its combinatorial meaning, making the connection between the triangle and counting explicit and widely known in the West.