Algebra and Trigonometry 10th Edition

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



Work Step by Step

$4·~_{n+1}P_2=~_{n+2}P_3$ $4~\frac{(n+1)!}{(n+1-2)!}=\frac{(n+2)!}{(n+2-3)!}$ $4=\frac{(n+2)!~(n+1-2)!}{(n+1)!~(n+2-3)!}=\frac{(n+2)(n+1)!~(n-1)!}{(n+1)!~(n-1)!}$ $4=n+2$ $n=2$
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