The Timpson Lyceum · Discrete Math (2025)

Binomial Coefficients and the Binomial Theorem - Lecture 9 (Discrete Math)

Overview

This lecture connects two ideas that look unrelated at first: the coefficients you get when you expand a binomial like $(x+y)^n$, and the number of ways to choose a subset of a given size from a set. We start by expanding small powers of $x+y$ by hand, notice the coefficient pattern in Pascal's triangle, and then introduce the "choose" notation $\binom{n}{k}$ for counting $k$-element subsets of an $n$-element set. We develop special cases and a general formula for $\binom{n}{k}$, prove the symmetry and Pascal identities, and finally state the Binomial Theorem. The payoff is that you can read off any single coefficient of an expansion — even something huge like $(x+y)^{17}$ — without multiplying everything out.

Key Concepts

Binomial. An expression with exactly two terms, e.g. $x+y$. Raising it to a power and collecting like terms produces a sum whose numerical multipliers are the binomial coefficients.

Small expansions. Multiplying out and collecting like terms gives a pattern where the exponent on $x$ counts down while the exponent on $y$ counts up:

• $(x+y)^1 = x + y$
• $(x+y)^2 = x^2 + 2xy + y^2$
• $(x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3$
• $(x+y)^4 = x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4$

Pascal's triangle. Start with a single $1$ (imagine zeros stretching off to both sides). Each entry in the next row is the sum of the two entries diagonally above it. Row $n$ (starting at $n=0$) lists the coefficients of $(x+y)^n$: $1;\quad 1\ 1;\quad 1\ 2\ 1;\quad 1\ 3\ 3\ 1;\quad 1\ 4\ 6\ 4\ 1;\quad 1\ 5\ 10\ 10\ 5\ 1;\ \dots$ The triangle is symmetric down its center.

Choose notation. For natural numbers $n,k$ (i.e. $0,1,2,\dots$), $\binom{n}{k}$ (read "$n$ choose $k$") is the number of $k$-element subsets of an $n$-element set — equivalently, the number of ways to choose $k$ things out of $n$ options. Order does not matter: choosing 3 of 10 people is the same as forming a 3-element subset.

Special cases. For any $n$:

• $\binom{n}{0} = 1$ (the empty set is the only 0-element subset).
• $\binom{n}{1} = n$.
• $\binom{n}{n} = 1$ (take everyone).
• $\binom{n}{n-1} = n$ (leave exactly one out).
• $\binom{n}{2} = 1 + 2 + 3 + \dots + (n-1) = \dfrac{n(n-1)}{2}$.

General formula. Choosing an ordered list of $k$ elements gives $n(n-1)(n-2)\cdots(n-k+1)$ possibilities; dividing by $k!$ to ignore order yields $\binom{n}{k} = \frac{n(n-1)(n-2)\cdots(n-k+1)}{k!} = \frac{n!}{(n-k)!\,k!}.$ Note the last factor in the falling product ends at $n-k+1$, not $n-k$.

Symmetry identity. $\binom{n}{k} = \binom{n}{n-k}$. Choosing $k$ to include is the same as choosing the $n-k$ to leave behind; taking complements pairs the $k$-subsets one-to-one with the $(n-k)$-subsets.

Pascal's identity. $\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$. Fix one special element: subsets either contain it (choose the remaining $k-1$ from the other $n-1$) or do not (choose all $k$ from the other $n-1$). Summing these disjoint cases counts every $k$-subset, which is exactly why each entry of Pascal's triangle is the sum of the two above it.

The connection to coefficients. Expanding $(x+y)^n$ without collecting terms produces one term per subset of an $n$-element set: read each factor as "$x$ = leave the element out, $y$ = include it." A term with $k$ copies of $y$ corresponds to a $k$-element subset, so after collecting like terms the coefficient of $x^{n-k}y^k$ is exactly the number of $k$-element subsets, $\binom{n}{k}$.

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

Worked Example

Find the coefficient of $x^{10}y^{7}$ in the expansion of $(x+y)^{17}$.

By the Binomial Theorem the coefficient of $x^{n-k}y^{k}$ is $\binom{n}{k}$. Here the exponent on $y$ is $k=7$ (and $n-k=10$ confirms $n=17$), so the coefficient is $\binom{17}{7}$:

Leaving the answer in this factored form is perfectly acceptable — it shows exactly how the count was built.

Summary

• The numbers multiplying each term of an expanded binomial $(x+y)^n$ are the binomial coefficients, and they appear as row $n$ of Pascal's triangle.
• $\binom{n}{k}$ ("$n$ choose $k$") counts the $k$-element subsets of an $n$-element set; order does not matter.
• Key special cases: $\binom{n}{0}=\binom{n}{n}=1$, $\binom{n}{1}=\binom{n}{n-1}=n$, and $\binom{n}{2}=\frac{n(n-1)}{2}$.
• The general formula is $\binom{n}{k}=\dfrac{n!}{(n-k)!\,k!}$, derived by counting ordered choices and dividing by $k!$.
• $\binom{n}{k}=\binom{n}{n-k}$ (symmetry) and $\binom{n}{k}=\binom{n-1}{k-1}+\binom{n-1}{k}$ (Pascal's identity, which explains the triangle).
• The Binomial Theorem $(x+y)^n=\sum_{k=0}^{n}\binom{n}{k}x^{n-k}y^{k}$ lets you read off any single coefficient directly, and the same counting idea solves problems like the number of non-backtracking lattice paths across a grid (a $10\times10$ grid gives $\binom{20}{10}$).