Algebra 2 (1st Edition)

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



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: $n!(n-3)=8n!\\n-3=8\\n=11$
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