Binomial Coefficient (n choose r): C(n,k) = n!/(k!(n−k)!) — Examples

Q: What is the difference between a binomial coefficient and a permutation?

A binomial coefficient $\binom{n}{r}$ counts the number of unordered selections (combinations) of $r$ items from $n$. A permutation $P(n, r) = \frac{n!}{(n-r)!}$ counts ordered arrangements. Since order doesn't matter in a combination, $\binom{n}{r}$ is always less than or equal to $P(n, r)$; specifically, $\binom{n}{r} = \frac{P(n,r)}{r!}$.

Q: Why is any number 'choose 0' equal to 1?

By the formula, $\binom{n}{0} = \frac{n!}{0!\cdot n!} = 1$ because $0! = 1$ by definition. Intuitively, there is exactly one way to choose nothing from a set: you simply select no items at all.

Binomial Coefficients

Numbers written in any of the ways shown below. Each notation is read aloud "n choose r."

A binomial coefficient equals the number of combinations of r items that can be selected from a set of n items. It also represents an entry in Pascal's triangle. These numbers are called binomial coefficients because they are coefficients in the binomial theorem.

Key Formula

\binom{n}{r} = \frac{n!}{r!\,(n-r)!}

Where:

$n$ = Total number of items in the set (a non-negative integer)
$r$ = Number of items being chosen, where 0 ≤ r ≤ n
$!$ = Factorial — the product of all positive integers up to that number (e.g., 5! = 120)

Worked Example

Problem: Compute the binomial coefficient

\binom{8}{3}

Step 1: Write out the formula with n = 8 and r = 3.

\binom{8}{3} = \frac{8!}{3!\,(8-3)!} = \frac{8!}{3!\cdot 5!}

Step 2: Expand only the numerator factors that don't cancel with 5! in the denominator. Since 8! = 8 × 7 × 6 × 5!, the 5! cancels.

\frac{8!}{3!\cdot 5!} = \frac{8 \times 7 \times 6}{3!}

Step 3: Evaluate 3! in the denominator.

3! = 3 \times 2 \times 1 = 6

Step 4: Divide to get the final result.

\frac{8 \times 7 \times 6}{6} = \frac{336}{6} = 56

Answer:

\binom{8}{3} = 56

. There are 56 ways to choose 3 items from a set of 8.

Another Example

This example applies the binomial coefficient to a real-world counting problem, reinforcing that 'n choose r' counts combinations. It also uses larger numbers than the first example, giving practice with the cancellation shortcut.

Problem: A pizza shop offers 10 toppings. You want to order a pizza with exactly 4 toppings. How many different pizzas can you create?

Step 1: Identify n and r. You have n = 10 toppings and are choosing r = 4.

\binom{10}{4} = \frac{10!}{4!\cdot 6!}

Step 2: Cancel 6! from the numerator and denominator. This leaves the top three factors of 10! above 6.

\frac{10!}{4!\cdot 6!} = \frac{10 \times 9 \times 8 \times 7}{4!}

Step 3: Compute 4! = 24 and multiply the numerator.

\frac{5040}{24} = 210

Answer: You can create 210 different 4-topping pizzas from 10 available toppings.

Frequently Asked Questions

What is the difference between a binomial coefficient and a permutation?

A binomial coefficient

\binom{n}{r}

counts the number of unordered selections (combinations) of

r

items from

n

. A permutation

P(n, r) = \frac{n!}{(n-r)!}

counts ordered arrangements. Since order doesn't matter in a combination,

\binom{n}{r}

is always less than or equal to

P(n, r)

; specifically,

\binom{n}{r} = \frac{P(n,r)}{r!}

Why is any number 'choose 0' equal to 1?

By the formula,

\binom{n}{0} = \frac{n!}{0!\cdot n!} = 1

because

0! = 1

by definition. Intuitively, there is exactly one way to choose nothing from a set: you simply select no items at all.

How are binomial coefficients related to Pascal's triangle?

Each entry in Pascal's triangle is a binomial coefficient. The entry in row

n

and position

r

(both starting from 0) equals

\binom{n}{r}

. The familiar addition rule of Pascal's triangle corresponds to the identity

\binom{n}{r} = \binom{n-1}{r-1} + \binom{n-1}{r}

Binomial Coefficient (Combination) vs. Permutation

	Binomial Coefficient (Combination)	Permutation
Definition	Number of unordered selections of r items from n	Number of ordered arrangements of r items from n
Formula	$\frac{n!}{r!(n-r)!}$	$\frac{n!}{(n-r)!}$
Order matters?	No — {A, B} and {B, A} count as the same selection	Yes — AB and BA count as different arrangements
Relationship	$\binom{n}{r} = \frac{P(n,r)}{r!}$	$P(n,r) = r! \cdot \binom{n}{r}$
Example (n=5, r=3)	$\binom{5}{3} = 10$	$P(5,3) = 60$

Why It Matters

Binomial coefficients appear throughout algebra, probability, and statistics. You use them whenever you need to count combinations—for instance, choosing a committee from a group, finding probabilities in binomial distributions, or expanding expressions like

(x + y)^n

. Mastering them is essential for courses from Algebra 2 through AP Statistics and beyond.

Common Mistakes

Mistake: Confusing the combination formula with the permutation formula by forgetting the

r!

in the denominator.

Correction: The combination formula divides by both

r!

and

(n-r)!

, while the permutation formula divides only by

(n-r)!

. Always check whether order matters: if it does not, you need the extra

r!

in the denominator.

Mistake: Computing

\binom{n}{r}

by fully expanding

n!

for large

n

, leading to enormous numbers and arithmetic errors.

Correction: Cancel common factors first. For

\binom{n}{r}

, write only the

r

factors

n \times (n-1) \times \cdots \times (n-r+1)

in the numerator and divide by

r!

. This keeps the numbers manageable.

Related Terms

Combination — Binomial coefficient counts combinations directly
Combination Formula — The formula used to compute binomial coefficients
Pascal's Triangle — Visual arrangement of all binomial coefficients
Binomial Theorem — Uses binomial coefficients to expand powers of binomials
Factorial — Key operation in the binomial coefficient formula
Permutation Formula — Counts ordered selections; related by dividing by r!
Binomial Coefficients in Pascal's Triangle — Shows how coefficients map to triangle entries
Set — Binomial coefficients count subsets of a given size