The Binomial Theorem and Pascal's Triangle

Main idea

When you multiply out (x + y)ⁿ, each term comes from picking x or y in every factor, so the coefficient of xⁿ⁻ᵏyᵏ is just the number of ways to pick which k factors give y: C(n, k). Stack those numbers row by row and you get Pascal's triangle.

(x + y)^{\htmlData{sym=0}{n}} = \sum_{{\htmlData{sym=1}{k}}=0}^{\htmlData{sym=0}{{n}}}{\htmlData{sym=2}{\binom{n}{k}}}x^{{\htmlData{sym=0}{n}}-{\htmlData{sym=1}{k}}}y^{\htmlData{sym=1}{k}}, \qquad {\htmlData{sym=2}{\binom{n}{k}}} = \binom{{\htmlData{sym=0}{n}}-1}{{\htmlData{sym=1}{k}}-1} + \binom{{\htmlData{sym=0}{n}}-1}{\htmlData{sym=1}{{k}}}

What the symbols mean

$n$
Power
How many factors of (x + y) are multiplied.
$k$
Term index
How many of the factors contribute y.
$\binom{n}{k}$
Binomial coefficient
The number of ways to pick those k factors.

Explanation

The binomial theorem is simply a shortcut for multiplying out $(x + y)^n$ without doing it by hand, and it turns out to be a counting problem in disguise. "Binomial" means "two terms," the $x$ and the $y$ . It's good for quick expansions and approximations, and it's the backbone of the binomial distribution in probability. The sum looks intimidating, but each piece is simple: $x^{n-k}y^k$ (one kind of term, with $k$ y's and the rest x's), $\binom{n}{k}$ (how many times that term shows up, which is how many ways to pick the $k$ factors that give $y$ ) and the $\sum$ (just add up all the kinds of terms, $k = 0$ to $n$ ). The second formula says each number in Pascal's triangle is the sum of the two above it. Stick with it; it's mostly the last lesson wearing algebra.

You've probably expanded $(x + y)^2 = x^2 + 2xy + y^2$ by hand. Where does that 2 come from? Think about what multiplying out actually does. In $(x + y)(x + y)$ you pick one term from each factor and multiply, in every possible way: $xx$ , $xy$ , $yx$ , $yy$ . The middle two are both $xy$ , so it shows up twice.

The same thing happens with more factors. Multiply out $(x + y)(x + y)(x + y)$ and every term comes from picking $x$ or $y$ in each of the three factors. You get $x^{3-k}y^k$ once for each way to choose which $k$ factors give $y$ , so its coefficient is $\binom{3}{k}$ from last lesson:

(x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3.

So expanding powers is secretly a counting problem.

Pascal's triangle lists $\binom{n}{k}$ row by row. Each entry is the sum of the two above it, and that's Pascal's identity $\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$ . It has a nice counting proof. To choose $k$ items from $n$ , look at item $n$ . Either you include it, and then you choose the other $k - 1$ from the remaining $n - 1$ , or you don't, and then all $k$ come from the remaining $n - 1$ . Those two cases don't overlap, so the sum rule adds them.

Row sums. Set $x = y = 1$ and the theorem gives $\sum_k \binom{n}{k} = (1 + 1)^n = 2^n$ . That makes sense from the counting side too: every subset of an $n$ -set has some size $k$ , so adding up the subsets of each size counts all $2^n$ subsets.

Coefficients with constants come from the same formula. In $(x + 2)^5$ , the $x^3$ term comes from picking $x$ in 3 factors and $2$ in the other 2, so it's $\binom{5}{2}x^3 \cdot 2^2 = 40x^3$ .

A common mistake. It's tempting to forget the constant's power and write the $x^3$ coefficient of $(x + 2)^5$ as just $\binom{5}{2} = 10$ . But each of those 10 picks multiplies in a $2$ twice, so every one contributes $4x^3$ , not $x^3$ .

See it

m

2.00

This is the first 32 rows of Pascal's triangle, with a dot wherever C(n, k) isn't divisible by m. With m = 2 the dots are the odd entries, and as you can see they form a Sierpiński triangle of nested gaps. It comes straight from Pascal's identity: odd + odd is even, so the pattern copies itself every 2, 4, 8, 16 rows.

Worked examples

Example 1

Find the coefficient of $x^3y^5$ in $(x + y)^8$ .

1
To get $x^3y^5$ you have to pick $y$ from 5 of the 8 factors and $x$ from the rest. So it's the $k = 5$ term, and its coefficient counts the ways to choose those 5 factors.

$\binom{8}{5} = 56$

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

What is the coefficient of $x^2y^3$ in $(x + y)^5$ ?

Problem 2+5 XP

What is the sum of the entries in row 10 of Pascal's triangle?

Problem 3+5 XP

What is the coefficient of $x^3$ in $(x + 2)^5$ ?

Problem 4+5 XP

Which identity says each entry of Pascal's triangle is the sum of the two entries above it?

Problem 5+5 XP

A basketball player makes 80% of free throws, independently. What's the probability she makes exactly 8 of her next 10?

Problem 6+5 XP

Use the first three terms of the binomial theorem to approximate $1.01^{10} = (1 + 0.01)^{10}$ .

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

Find $\displaystyle\sum_{k=0}^{10} k\binom{10}{k}$ .

Where it's used

The probability of exactly k successes in n independent trials (the binomial distribution) uses these coefficients, from quality control to genetics.

Permutations and Combinations Counting Multisets: Stars and Bars