Answer
$n=11$
Work Step by Step
$_nP_r=\frac{n!}{(n-r)!}$, hence here the equation is: $\frac{n!}{(n-4)!}=8\frac{n!}{(n-3)!}$
If we multiply both sides by $(n-3)!=(n-3)\cdot(n-4)!$ we get: $n!(n-3)=8n!\\n-3=8\\n=11$
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: 58
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