## Thinking Mathematically (6th Edition)

Published by Pearson

# Chapter 11 - Counting Methods and Probability Theory - 11.3 Combinations - Exercise Set 11.3 - Page 707: 35

#### Answer

$22,957,480$ selections

#### Work Step by Step

A combination from a group of items occurs when no item is used more than once and the order of items makes no difference. The number of combinations possible if $r$ items are taken from $n$ items is ${}_{n}C_{r}=\displaystyle \frac{n!}{(n-r)!r!}$ -------------- Order is not important, we deal with combinations. ${}_{53}C_{6}=\displaystyle \frac{53!}{(53-6)!6!}$ $=\displaystyle \frac{53\times 52\times 51\times 50\times 49\times 48}{1\times 2\times 3\times 4\times 5\times 6}$ $=\displaystyle \frac{53\times 52\times 17\times 10\times 49\times 1}{1\times 1\times 1\times 1\times 1\times 1}$ $=22,957,480$

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