Chapter 11 - Counting Methods and Probability Theory - Chapter Summary, Review, and Test - Review Exercises - Page 758: 13

$330$

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: ${}_{n}C_{r}=\displaystyle \frac{n!}{(n-r)!r!}$ --- ${}_{11}C_{7}=\displaystyle \frac{11!}{(11-7)!7!}=\frac{11!}{4!\cdot 7!}\qquad$... 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) $=11\times 10\times 3$ $=330$

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