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Combination

Combination

A selection of objects from a collection. Order is irrelevant.

 

Example: A poker hand is a combination of 5 cards from a 52 card deck. This is a combination since the order of the 5 cards does not matter.

 

See also

Binomial coefficient, combination formula, permutation

Key Formula

C(n,r)=(nr)=n!r!(n−r)!C(n, r) = \binom{n}{r} = \frac{n!}{r!(n - r)!}
Where:
  • nn = The total number of items in the collection
  • rr = The number of items you are selecting
  • !! = Factorial — the product of all positive integers up to that number (e.g., 4! = 4 × 3 × 2 × 1 = 24)

Worked Example

Problem: A pizza shop offers 8 toppings. You can choose any 3 toppings for your pizza. How many different 3-topping pizzas are possible?
Step 1: Identify n and r. There are 8 toppings total (n = 8), and you are choosing 3 of them (r = 3). Order does not matter — picking pepperoni, mushrooms, olives is the same pizza as olives, pepperoni, mushrooms.
n=8,r=3n = 8, \quad r = 3
Step 2: Apply the combination formula.
(83)=8!3!(8−3)!=8!3!⋅5!\binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8!}{3! \cdot 5!}
Step 3: Simplify. Instead of computing the full factorials, cancel the 5! from numerator and denominator.
8×7×63×2×1=3366=56\frac{8 \times 7 \times 6}{3 \times 2 \times 1} = \frac{336}{6} = 56
Answer: There are 56 different 3-topping pizzas you can make from 8 toppings.

Another Example

Problem: A committee of 4 people must be chosen from a group of 10 volunteers. How many different committees are possible?
Step 1: Recognize this is a combination problem because the committee has no ranked positions — just 4 members chosen from 10.
n=10,r=4n = 10, \quad r = 4
Step 2: Apply the formula and simplify by canceling 6! from numerator and denominator.
(104)=10!4!⋅6!=10×9×8×74×3×2×1\binom{10}{4} = \frac{10!}{4! \cdot 6!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1}
Step 3: Compute the arithmetic.
504024=210\frac{5040}{24} = 210
Answer: There are 210 possible committees of 4 from 10 volunteers.

Frequently Asked Questions

How do I know when to use a combination versus a permutation?
Ask yourself: does rearranging the selected items create a different outcome? If yes, use a permutation (order matters). If no — like choosing team members, lottery numbers, or pizza toppings — use a combination (order does not matter).
Why do you divide by r! in the combination formula?
The permutation formula counts every possible ordering of the selected items as a separate arrangement. Dividing by r! removes those duplicate orderings. For instance, if you select 3 items, those 3 items can be arranged in 3! = 6 ways, so you divide by 6 to count the group only once.

Combination vs. Permutation

Both involve selecting items from a group. A combination counts selections where order does not matter, while a permutation counts arrangements where order does matter. Choosing 3 books to take on vacation is a combination; ranking your top 3 books from favorite to least favorite is a permutation. The permutation formula is P(n,r)=n!(n−r)!P(n,r) = \frac{n!}{(n-r)!}, which is always greater than or equal to the corresponding combination C(n,r)=n!r!(n−r)!C(n,r) = \frac{n!}{r!(n-r)!} because it counts each group multiple times — once for every ordering.

Why It Matters

Combinations are fundamental to probability and statistics. Whenever you need to count equally likely outcomes — such as lottery draws, card hands, or survey samples — you use combinations to find how many ways events can occur. They also appear throughout algebra as binomial coefficients, which are the numbers in Pascal's triangle and the coefficients in the expansion of (a+b)n(a + b)^n.

Common Mistakes

Mistake: Using a permutation formula when order does not matter, which overcounts the result.
Correction: Always ask whether rearranging your selection produces a genuinely different outcome. If it doesn't, divide by r! (use the combination formula) to remove duplicate orderings.
Mistake: Confusing n and r — putting the number chosen on top and the total on the bottom.
Correction: Remember that n (the larger number, the total) always goes on top in the formula. You choose r items from n, so (nr)\binom{n}{r} has n ≥ r.

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