Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 13 - Counting and Probability - 13.2 Permutations and Combinations - 13.2 Asses Your Understanding - Page 855: 38



Work Step by Step

The order of the letters matters here, so the situation involves permutation. We know that $P(n,r)=\frac{n!}{(n-r)!}.$ Hence with $n=6$, $r=4$, then: $P(6,4)=\frac{6!}{(6-4)!}=\frac{6\cdot5\cdot4\cdot3\cdot2!}{2!}=6\cdot5\cdot4\cdot3=360.$
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