Mathwords logoReference LibraryMathwords

Combination Formula

Combination Formula

A formula for the number of possible combinations of r objects from a set of n objects. This is written in any of the ways shown below.

Three equivalent notations for binomial coefficients: (n choose r), nCr, C(n,r), and C with superscript n and subscript r.

All forms are read aloud "n choose r."

 

Formula: Formula: C(n,r) or nCr = n! / (r!(n-r)!) = n(n-1)(n-2)···(n-r+1) / r!
  Note: Formula showing binomial coefficient: (n choose r) equals n P r divided by r factorial, where nPr is the formula for permutations of n objects taken r at a time.
Example:

How many different committees of 4 students can be chosen from a group of 15?
Answer: There are The binomial coefficient notation "15 choose 4," written as C(15,4) in parenthetical form. possible combinations of 4 students from a set of 15.
 

C(15,4) = 15!/(4!11!) = (15·14·13·12)/(4·3·2·1) = 1365

There are 1365 different committees.

 

See also

Permutation, binomial coefficient, factorial

Key Formula

(nr)=C(n,r)=n!r! (n−r)!\binom{n}{r} = C(n,r) = \frac{n!}{r!\,(n-r)!}
Where:
  • nn = The total number of objects in the set
  • rr = The number of objects being chosen
  • !! = Factorial — the product of all positive integers up to that number (e.g., 5! = 5 × 4 × 3 × 2 × 1 = 120)

Worked Example

Problem: How many different committees of 4 students can be chosen from a group of 15?
Step 1: Identify n and r. You have 15 students total and want to choose 4, so n = 15 and r = 4.
(154)=15!4! (15−4)!\binom{15}{4} = \frac{15!}{4!\,(15-4)!}
Step 2: Simplify the denominator. Since (15 − 4)! = 11!, most factors in 15! cancel with 11!.
15!4!⋅11!=15×14×13×124!\frac{15!}{4! \cdot 11!} = \frac{15 \times 14 \times 13 \times 12}{4!}
Step 3: Calculate the numerator.
15×14×13×12=32,76015 \times 14 \times 13 \times 12 = 32{,}760
Step 4: Calculate the denominator. 4! = 4 × 3 × 2 × 1 = 24.
4!=244! = 24
Step 5: Divide to get the final answer.
32,76024=1,365\frac{32{,}760}{24} = 1{,}365
Answer: There are 1,365 different committees of 4 students that can be chosen from a group of 15.

Another Example

This example uses smaller numbers and r = 2, making the factorial cancellation simpler and showing how quickly the formula resolves for small selections.

Problem: A pizza shop offers 8 toppings. How many ways can you choose 2 toppings for your pizza?
Step 1: Identify n and r. There are 8 toppings and you choose 2, so n = 8 and r = 2.
(82)=8!2! (8−2)!\binom{8}{2} = \frac{8!}{2!\,(8-2)!}
Step 2: Simplify by canceling 6! from the numerator and denominator.
8!2!⋅6!=8×72!\frac{8!}{2! \cdot 6!} = \frac{8 \times 7}{2!}
Step 3: Compute the numerator and denominator, then divide.
562=28\frac{56}{2} = 28
Answer: There are 28 ways to choose 2 toppings from 8 options.

Frequently Asked Questions

What is the difference between a combination and a permutation?
A combination counts selections where order does not matter, while a permutation counts arrangements where order does matter. For example, choosing players A, B, and C for a team is one combination but six different permutations (ABC, ACB, BAC, BCA, CAB, CBA). The combination formula divides the permutation formula by r! to remove duplicate orderings.
When do you use the combination formula?
Use the combination formula whenever you are selecting a group from a larger set and the order of selection is irrelevant. Common scenarios include choosing committee members, picking lottery numbers, selecting pizza toppings, or forming teams. If the order matters — such as assigning 1st, 2nd, and 3rd place — use the permutation formula instead.
Why does C(n, r) equal C(n, n − r)?
Every time you choose r objects to include, you are simultaneously choosing (n − r) objects to exclude. This symmetry means the number of ways to pick r items is identical to the number of ways to pick (n − r) items. For instance, C(10, 3) = C(10, 7) = 120. This property is useful for simplifying calculations when r is close to n.

Combination Formula vs. Permutation Formula

Combination FormulaPermutation Formula
DefinitionNumber of ways to choose r objects from n, order does NOT matterNumber of ways to arrange r objects from n, order DOES matter
Formulan! / (r! · (n − r)!)n! / (n − r)!
RelationshipC(n, r) = P(n, r) / r!P(n, r) = C(n, r) × r!
Example: 5 items, choose 3C(5, 3) = 10P(5, 3) = 60
When to useSelecting committees, lottery draws, subsetsRanking, seating arrangements, assigning positions

Why It Matters

The combination formula appears throughout algebra, probability, and statistics courses. It is essential for computing probabilities of events like drawing cards from a deck or selecting lottery numbers. It also forms the basis of the binomial coefficients used in the Binomial Theorem for expanding expressions like (a+b)n(a + b)^n.

Common Mistakes

Mistake: Using the permutation formula when order doesn't matter, producing an answer that is r! times too large.
Correction: Ask yourself: does the order of selection matter? If choosing a committee of 3 people, the group {A, B, C} is the same regardless of the order picked. Use the combination formula and divide by r! to eliminate duplicate orderings.
Mistake: Forgetting to simplify factorials before multiplying, leading to unnecessarily large numbers or calculator overflow.
Correction: Cancel common factors first. For C(15, 4), write (15 × 14 × 13 × 12) / (4 × 3 × 2 × 1) instead of computing 15! and 11! separately. You can also divide stepwise: 15/1 × 14/2 × 13/3 × 12/4 to keep numbers manageable.

Related Terms