Answer
The identity is verified.
$_nP_{n-1}=~_nP_n$
Work Step by Step
$_nP_{n-1}=\frac{n!}{[(n-(n-1)]!}=\frac{n!}{1!}$
But, $1!=1=0!$
$_nP_{n-1}=\frac{n!}{1!}=\frac{n!}{0!}=\frac{n!}{(n-n)!}=~_nP_n$
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: 87
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
