Mathwords logoMathwords

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

Binomial Coefficients

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

(nr)   or   nCr   or   C(n,r)   or occasionally   Crn\binom{n}{r} \;\text{ or }\; {}_nC_r \;\text{ or }\; C(n,r) \;\text{ or occasionally }\; \mathrm{C}_r^n

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.

Formula: (nr) or nCr=n!r! (n−r)!=n (n−1) (n−2)⋯(n−r+1)r!\begin{aligned} \binom{n}{r} \text{ or } {}_{n}C_{r} &= \frac{n!}{r!\,(n-r)!} \\ &= \frac{n\,(n-1)\,(n-2)\cdots(n-r+1)}{r!} \end{aligned}
  Note: (nr)=nPrr!,\displaystyle \binom{n}{r} = \frac{{}_{n}P_{r}}{r!}\text{,} where nPr is the formula for permutations of n objects taken r at a time.
Examples:

(72)=7!2!5!=7⋅62⋅1=21\binom{7}{2} = \frac{7!}{2!5!} = \frac{7 \cdot 6}{2 \cdot 1} = 21
 

(75)=7!5!2!=7⋅6⋅5⋅4⋅35⋅4⋅3⋅2⋅1=7⋅62⋅1=21\binom{7}{5} = \frac{7!}{5!2!} = \frac{7 \cdot 6 \cdot 5 \cdot 4 \cdot 3}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = \frac{7 \cdot 6}{2 \cdot 1} = 21

20C3=20!3!17!=20⋅19⋅183⋅2⋅1=1140{}_{20}C_{3} = \frac{20!}{3!17!} = \frac{20 \cdot 19 \cdot 18}{3 \cdot 2 \cdot 1} = 1140

See also

Factorial, binomial coefficients in Pascal's triangle, combination formula, permutations

Key Formula

(nr)=n!r! (n−r)!\binom{n}{r} = \frac{n!}{r!\,(n-r)!}
Where:
  • nn = Total number of items in the set (a non-negative integer)
  • rr = 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 (83)\binom{8}{3}.
Step 1: Write out the formula with n = 8 and r = 3.
(83)=8!3! (8−3)!=8!3!⋅5!\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.
8!3!⋅5!=8×7×63!\frac{8!}{3!\cdot 5!} = \frac{8 \times 7 \times 6}{3!}
Step 3: Evaluate 3! in the denominator.
3!=3×2×1=63! = 3 \times 2 \times 1 = 6
Step 4: Divide to get the final result.
8×7×66=3366=56\frac{8 \times 7 \times 6}{6} = \frac{336}{6} = 56
Answer: (83)=56\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.
(104)=10!4!⋅6!\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.
10!4!⋅6!=10×9×8×74!\frac{10!}{4!\cdot 6!} = \frac{10 \times 9 \times 8 \times 7}{4!}
Step 3: Compute 4! = 24 and multiply the numerator.
504024=210\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 (nr)\binom{n}{r} counts the number of unordered selections (combinations) of rr items from nn. A permutation P(n,r)=n!(n−r)!P(n, r) = \frac{n!}{(n-r)!} counts ordered arrangements. Since order doesn't matter in a combination, (nr)\binom{n}{r} is always less than or equal to P(n,r)P(n, r); specifically, (nr)=P(n,r)r!\binom{n}{r} = \frac{P(n,r)}{r!}.
Why is any number 'choose 0' equal to 1?
By the formula, (n0)=n!0!⋅n!=1\binom{n}{0} = \frac{n!}{0!\cdot n!} = 1 because 0!=10! = 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 nn and position rr (both starting from 0) equals (nr)\binom{n}{r}. The familiar addition rule of Pascal's triangle corresponds to the identity (nr)=(n−1r−1)+(n−1r)\binom{n}{r} = \binom{n-1}{r-1} + \binom{n-1}{r}.

Binomial Coefficient (Combination) vs. Permutation

Binomial Coefficient (Combination)Permutation
DefinitionNumber of unordered selections of r items from nNumber of ordered arrangements of r items from n
Formulan!r!(n−r)!\frac{n!}{r!(n-r)!}n!(n−r)!\frac{n!}{(n-r)!}
Order matters?No — {A, B} and {B, A} count as the same selectionYes — AB and BA count as different arrangements
Relationship(nr)=P(n,r)r!\binom{n}{r} = \frac{P(n,r)}{r!}P(n,r)=r!⋅(nr)P(n,r) = r! \cdot \binom{n}{r}
Example (n=5, r=3)(53)=10\binom{5}{3} = 10P(5,3)=60P(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(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!r! in the denominator.
Correction: The combination formula divides by both r!r! and (n−r)!(n-r)!, while the permutation formula divides only by (n−r)!(n-r)!. Always check whether order matters: if it does not, you need the extra r!r! in the denominator.
Mistake: Computing (nr)\binom{n}{r} by fully expanding n!n! for large nn, leading to enormous numbers and arithmetic errors.
Correction: Cancel common factors first. For (nr)\binom{n}{r}, write only the rr factors n×(n−1)×⋯×(n−r+1)n \times (n-1) \times \cdots \times (n-r+1) in the numerator and divide by r!r!. This keeps the numbers manageable.

Related Terms