## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 10 - Section 10.6 - Counting Principles, Permutations, and Combinations - Exercise Set - Page 1103: 6

#### Answer

The required solution is $362880$

#### Work Step by Step

We know that the representation $_{n}{{P}_{r}}$ implies that the number of possible well-organized arrangements of n items is taken r at a time. And the number of possible well-organized arrangements of n items taken r at a time can be evaluated as: $_{n}{{P}_{r}}=\frac{n!}{\left( n-r \right)!}$ And the provided expression is $_{9}{{P}_{9}}$. Here, $n=9,r=9$. Put the value of n, r in the above formula. Then: \begin{align} & _{9}{{P}_{9}}=\frac{9!}{\left( 9-9 \right)!} \\ & =\frac{9!}{0!} \\ & =\frac{9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{0!} \end{align} Since, $0!=1$. Therefore, \begin{align} & \frac{9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{0!}=\frac{9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{1} \\ & =\frac{362880}{1} \\ & =362880 \end{align} Thus, $_{9}{{P}_{9}}=362880$

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