Answer
$n=2$
Work Step by Step
$4·~_{n+1}P_2=~_{n+2}P_3$
$4~\frac{(n+1)!}{(n+1-2)!}=\frac{(n+2)!}{(n+2-3)!}$
$4=\frac{(n+2)!~(n+1-2)!}{(n+1)!~(n+2-3)!}=\frac{(n+2)(n+1)!~(n-1)!}{(n+1)!~(n-1)!}$
$4=n+2$
$n=2$
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: 75
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