Chapter 7 - Section 7.6 - Counting Theory - 7.6 Exercises - Page 678: 10

$120$

Using $P(n,r)=\dfrac{n!}{(n-r)!},$ the given expression, $P( 5,5 ),$ evaluates to \begin{array}{l}\require{cancel} =\dfrac{5!}{(5-5)!} \\\\= \dfrac{5!}{0!} \\\\= \dfrac{5(4)(3)(2)(1)}{1} \\\\= \dfrac{120}{1} \\\\= 120 \end{array}