#### Answer

56

#### Work Step by Step

We know that:
$_nC_r =\frac{_nP_r}{r!} = \frac{n!}{(n-r)!r!}$
$_8C_5 = \frac{8!}{(8-5)!5!}$
$\frac{ 8!}{5!3!}$
We also know that:
$x! = x(x-1)(x-2)...(1)$
Thus, we have:
56

Chegg costs money, GradeSaver solutions are free!

Published by
Prentice Hall

ISBN 10:
0133281140

ISBN 13:
978-0-13328-114-9

56

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.