Chapter 11 - Counting Methods and Probability Theory - Chapter Summary, Review, and Test - Review Exercises - Page 758: 10

$990$

Factorial Notation $n!=n(n-1)(n-2)\cdots(3)(2)(1) $ and $0!=1 $(by definition) It follows that $n!=n(n-1)!$ --- $11!=11(10!)=11(10)(9!)=11(10)(9)(8!)$ $(11-3)!=8!$ $\displaystyle \frac{11!}{(11-3)!}=\frac{11(10)(9)(8!)}{8!}\qquad$ ... reduce the fraction (divide with $\displaystyle \frac{8!}{8!}$) $=11(10)(9)=990$