Answer
21
Work Step by Step
We know that:
$_nC_r =\frac{_nP_r}{r!} = \frac{n!}{(n-r)!r!}$
$_7C_5 = \frac{7!}{(7-5)!5!}$
$\frac{ 7!}{5!2!}$
We also know that:
$x! = x(x-1)(x-2)...(1)$
Thus, we have:
21
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