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 - Exercise Set - Page 1105: 89

Answer

The statement is true.

Work Step by Step

Let us consider the left hand side expression of the provided equation, $_{7}{{P}_{3}}$ Recall, $_{n}{{\operatorname{P}}_{r}}=\frac{n!}{\left( n-r \right)!}$ Therefore, $\begin{align} & _{7}{{P}_{3}}=\frac{7!}{\left( 7-3 \right)!} \\ & =\frac{7!}{4!} \\ & =\frac{7\times 6\times 5\times 4!}{4!} \\ & =210 \end{align}$ Also, consider the left hand side expression of the provided equation, $3!\cdot \left( _{7}{{C}_{3}} \right)$ Recall, $_{n}{{C}_{r}}=\frac{n!}{r!\left( n-r \right)!}$ Then, $\begin{align} & 3!\cdot \left( _{7}{{C}_{3}} \right)=3!\times \frac{7!}{3!\left( 7-3 \right)!} \\ & =3!\times \frac{7!}{3!\left( 4 \right)!} \\ & =\frac{7!}{\left( 4 \right)!} \\ & =\frac{7\times 6\times 5\times 4!}{4!} \end{align}$ So, $3!\cdot \left( _{7}{{C}_{3}} \right)=210$ Thus, $_{7}{{P}_{3}}=3!\cdot \left( _{7}{{C}_{3}} \right)$
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