Answer
$330$
Work Step by Step
See review summary:
11.1.b. Factorial Notation
$n!=n(n-1)(n-2)\cdots(3)(2)(1) $ and $0!=1 $(by definition)
( It follows that $n!=n(n-1)!$ )
11.3.b. Combinations Formula: ${}_{n}C_{r}=\displaystyle \frac{n!}{(n-r)!r!}$
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${}_{11}C_{7}=\displaystyle \frac{11!}{(11-7)!7!}=\frac{11!}{4!\cdot 7!}\qquad $... apply $n!=n(n-1)!$
$=\displaystyle \frac{11\times 10\times 9\times 8\times 7!}{4!\cdot 7!}\qquad $... reduce the fraction (divide with $\displaystyle \frac{7!}{7!}$)
$=\displaystyle \frac{11\times 10\times[9]\times(8)}{(4)\times[3]\times(2)\times 1}\qquad $... reduce the fraction (with 4,2,3)
$=11\times 10\times 3$
$=330$
