College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 7 - Section 7.6 - Counting Theory - 7.6 Exercises - Page 678: 13

Answer

$5$

Work Step by Step

Using $P(n,r)=\dfrac{n!}{(n-r)!},$ the given expression, $P( 5,1 ),$ evaluates to \begin{array}{l}\require{cancel} =\dfrac{5!}{(5-1)!} \\\\= \dfrac{5!}{4!} \\\\= \dfrac{5(4!)}{4!} \\\\= \dfrac{5(\cancel{4!})}{\cancel{4!}} \\\\= 5 \end{array}
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