# Chapter 13 - Probability - 13-3 Permutations and Combinations - Practice and Problem-Solving Exercises - Page 841: 15

210

#### Work Step by Step

We know that: $_nC_r =\frac{_nP_r}{r!} = \frac{n!}{(n-r)!r!}$ $_{10}C_6 = \frac{10!}{(10-6)!6!}$ $\frac{ 10!}{6!4!}$ We also know that: $x! = x(x-1)(x-2)...(1)$ Thus, we have: 210

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