Thinking Mathematically (6th Edition)

Published by Pearson
ISBN 10: 0321867327
ISBN 13: 978-0-32186-732-2

Chapter 11 - Counting Methods and Probability Theory - 11.2 Permutations - Exercise Set 11.2 - Page 701: 38

Answer

362880

Work Step by Step

The number of possible permutations if r items are taken from n items is nPr = $\frac{n!}{(n-r)!}$ 9P9= $\frac{9!}{(9-9)!}$ = $\frac{9!}{0!}$ = $\frac{362880}{1}$ = 362880 We put 1 on the bottom since 0!=1.
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