Combination Math Example 4

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

Example 4

medium
A pizza shop offers 88 toppings. How many different 33-topping pizzas can be made if order does not matter?

Solution

  1. 1
    Because order does not matter, use combinations: (83)=8!3!5!\binom{8}{3} = \frac{8!}{3!5!}.
  2. 2
    Simplify: (83)=8×7×63×2×1=56\binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56.

Answer

5656
Combinations count selections when only the chosen group matters. A pizza with pepperoni, mushroom, and onion is the same pizza no matter what order the toppings are listed.

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 1 easy

A committee of [formula] is to be chosen from [formula] people. How many different committees are po

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?

← All Combination Examples Combination Concept