Answer
$ 2002$
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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${}_{14}C_{5}=\displaystyle \frac{14!}{(14-5)!5!}=\frac{14!}{9!\cdot 5!}\qquad $... apply $n!=n(n-1)!$
$=\displaystyle \frac{14\times 13\times 12\times 11\times 10\times 9!}{9!\cdot 5!}\qquad $... reduce the fraction (divide with $\displaystyle \frac{9!}{9!}$)
$=\displaystyle \frac{14\times 13\times(12)\times 11\times[10]}{[5]\times(4\times 3)\times[2]\times 1}\qquad $... reduce the fraction (with 5,4,3,2)
$=14\times 13\times 1\times 11\times 1$
$=2002$
