Thinking Mathematically (6th Edition)

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

Chapter 11 - Counting Methods and Probability Theory - 11.2 Permutations - Exercise Set 11.2: 63

Answer

12

Work Step by Step

A permutation from a group of items occurs when no item is used more than once and the order of arrangement makes a difference. The number of permutations possible if $r$ items are taken from $n$ items is ${}_{n}P_{r}=\displaystyle \frac{n!}{(n-r)!}$. The number of permutations of $n$ items, where $p$ items are identical, $q$ items are identical, $r$ items are identical, and so on, is $\displaystyle \frac{n!}{p!q!r!\ldots}$ ------------------------ BAKE has four letters with no duplicates, so the number of permutations is ${}_{4}P_{4}=4!=24.$ BABE has four letters, among which there are duplicates, 2 letters B, ... p=2, and the number of distinct permutations is $\displaystyle \frac{4!}{2!}=\frac{24}{2}=12$
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