# Chapter 13 - Probability - 13-3 Permutations and Combinations - Got It? - Page 839: 4

56

#### Work Step by Step

This is given by: $_{8}C_5$ We know that: $_nC_r =\frac{_nP_r}{r!} = \frac{n!}{(n-r)!r!}$ $_{8}C_5 = \frac{8!}{(8-5)!5!}$ We also know that: $x! = x(x-1)(x-2)...(1)$ Thus, we have: 56

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