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: 12


$P(9,0) = 1$

Work Step by Step

With $P(n,r)=\dfrac{n!}{(n-r)!}$: then, $P(9,0) = \dfrac{9!}{(9-0)!} = \dfrac{9!}{9!} =1$
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