Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 10 - Section 10.6 - Counting Principles, Permutations, and Combinations - Concept and Vocabulary Check - Page 1103: 5

Answer

The formula for $_{n}{{C}_{r}}$ has the same numerator as the formula for $_{n}{{P}_{r}}$ but contains an extra factor of $r!$ in the denominator.

Work Step by Step

We know that permutations of $n$ things taken $r$ at a time can be defined by the formula: ${}_{n}{{P}_{r}}=\frac{n!}{\left( n-r \right)!}$ And combinations of $n$ things taken $r$ at a time can be defined by the formula: ${}_{n}{{C}_{r}}=\frac{n!}{r!\left( n-r \right)!}$ And by comparing both formulas it can be seen that r! is an extra factor. Therefore, there are r! times as many permutations of n things taken r at a time as combinations of n things taken r at a time.
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