Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 12 - Counting and Probability - Section 12.2 Permutations and Combinations - 12.2 Assess Your Understanding - Page 875: 25

Answer

$P(4, 3)=24$ Refer to the list below.

Work Step by Step

If we want to choose $r$ elements out of $n$ regarding the order, not allowing repetition, we can do this in $P(n, r)=\frac{n!}{(n-r)!}$ ways. Hence, $P(4, 3)=\dfrac{4!}{(4-3)!}=\dfrac{4!}{1!}=4\cdot3\cdot2\cdot1=24$ The list of all arrangements: $123,124,132,134,142,143\\ 213,214,231,234,241,243\\ 312,314,321,324,341,342\\ 412,413,421,423,431,432$
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