Thinking Mathematically (6th Edition)

Published by Pearson
ISBN 10: 0321867327
ISBN 13: 978-0-32186-732-2

Chapter 11 - Counting Methods and Probability Theory - 11.3 Combinations - Exercise Set 11.3: 31

Answer

$495$

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 ${}_{n}C_{r}=\displaystyle \frac{n!}{(n-r)!r!}$ -------------- It does not matter which book was chosen first, second,... Order is not important, we deal with combinations. ${}_{12}C_{4}=\displaystyle \frac{13!}{(12-4)!4!}=\frac{12\times 11\times 10\times 9}{1\times 2\times 3\times 4}$ =$11\times 5\times 9=495$
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