## Thinking Mathematically (6th Edition)

$2002$
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: ${}_{n}C_{r}=\displaystyle \frac{n!}{(n-r)!r!}$ --- ${}_{14}C_{5}=\displaystyle \frac{14!}{(14-5)!5!}=\frac{14!}{9!\cdot 5!}\qquad$... 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) $=14\times 13\times 1\times 11\times 1$ $=2002$