Answer
$n=13$
Work Step by Step
$_nP_r=\frac{n!}{(n-r)!}$, hence here the equation is: $\frac{n!}{(n-5)!}=9\frac{n!}{(n-4)!}$
If we multiply both sides by $(n-4)!=(n-4)\cdot(n-5)!$ we get: $n!(n-4)=9n!\\n-4=9\\n=13$
Algebra 2 (1st Edition)
by Larson, Ron; Boswell, Laurie; Kanold, Timothy D.; Stiff, Lee
Published by
McDougal Littell
ISBN 10:
0618595414
ISBN 13:
978-0-61859-541-9
Chapter 10 Counting Methods and Probability - 10.1 Apply the counting Principles and Permutations - 10.1 Exercises - Skill Practice - Page 687: 60
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