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

$151,200$

#### Work Step by Step

See review summary: b. Factorial Notation $n!=n(n-1)(n-2)\cdots(3)(2)(1)$ and $0!=1$(by definition) ( It follows that $n!=n(n-1)!$ ) c. Permutations Formula$\displaystyle \quad {}_{n}P_{r}=\frac{n!}{(n-r)!}$ --- ${}_{10}P_{6}=\displaystyle \frac{10!}{(10-6)!}=\frac{10!}{4!}\qquad$... apply $n!=n(n-1)!$ $=\displaystyle \frac{10\times 9!}{4!}$ $=\displaystyle \frac{10\times 9\times 8\times 7\times 6\times 5\times 4!}{4!}\qquad$ ... reduce the fraction (divide with $\displaystyle \frac{4!}{4!}$) =$10\times 9\times 8\times 7\times 6\times 5$ = $151,200$

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