Answer
$n=3$
Work Step by Step
$_{n+1}P_3=4·~_{n}P_2$
$\frac{(n+1)!}{(n+1-3)!}=4~\frac{n!}{(n-2)!}$
$\frac{(n+1)!~(n-2)!}{n!~(n+1-3)!}=4$
$\frac{(n+1)n!~(n-2)!}{n!~(n-2)!}=4$
$n+1=4$
$n=3$
Algebra and Trigonometry 10th Edition
by Larson, Ron
Published by
Cengage Learning
ISBN 10:
9781337271172
ISBN 13:
978-1-33727-117-2
Chapter 11 - 11.6 - Counting Principles - 11.6 Exercises - Page 825: 77
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