Chapter 13 - Probability - 13-3 Permutations and Combinations - Practice and Problem-Solving Exercises - Page 841: 18

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We use the combination equation since order does not matter. We know that: $_nC_r =\frac{_nP_r}{r!} = \frac{n!}{(n-r)!r!}$ $_7C_2 = \frac{7!}{(7-2)!2!}$ $\frac{ 7!}{5!2!}$ We also know that: $x! = x(x-1)(x-2)...(1)$ Thus, we have: 21