Combination Math Example 1

Follow the full solution, then compare it with the other examples linked below.

Example 1

easy
A committee of 33 is to be chosen from 88 people. How many different committees are possible?

Solution

  1. 1
    Recall the combination formula for unordered selections: (nr)=n!r!(n−r)!\binom{n}{r} = \frac{n!}{r!(n-r)!}, with n=8n = 8, r=3r = 3.
  2. 2
    Cancel common factorial terms: (83)=8!3!⋅5!=8×7×63×2×1\binom{8}{3} = \frac{8!}{3! \cdot 5!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1}
  3. 3
    Calculate: 3366=56\frac{336}{6} = 56

Answer

(83)=56\binom{8}{3} = 56
Combinations count selections where order does not matter. Choosing members A, B, C for a committee is the same as choosing C, B, A.

About Combination

A combination is an unordered selection of objects — the number of ways to choose rr items from nn distinct items is C(n,r)=n!r!(n−r)!C(n,r) = \frac{n!}{r!(n-r)!}.

Learn more about Combination →

More Combination Examples

Example 2 medium

From a group of [formula] men and [formula] women, how many committees of [formula] can be formed th

Example 3 medium

How many ways can you choose [formula] books from a shelf of [formula] books?

Example 4 medium

A pizza shop offers [formula] toppings. How many different [formula]-topping pizzas can be made if o

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