Chapter 7 - Section 7.6 - Counting Theory - 7.6 Exercises - Page 679: 22

$9,034,502,400$

Using $_nP_r=\dfrac{n!}{(n-r)!},$ the given expression, $ _{100}P_{5} ,$ evaluates to \begin{array}{l}\require{cancel} =\dfrac{100!}{(100-5)!} \\\\= \dfrac{100!}{95!} \\\\= \dfrac{100(99)(98)(97)(96)(\cancel{95!})}{\cancel{95!}} \\\\= 9,034,502,400 \end{array}