## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 10 - Section 10.6 - Counting Principles, Permutations, and Combinations - Exercise Set - Page 1105: 80

#### Answer

The required solution is ${}_{n}{{C}_{r}}=\frac{n!}{r!\left( n-r \right)!}$

#### Work Step by Step

We know that the above problem uses combinations as the order of the selection does not matter. The formula is given by: ${}_{n}{{C}_{r}}=\frac{n!}{r!\left( n-r \right)!}$

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