Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 11 - Further Topics in Algebra - Chapter 11 Test Prep - Review Exercises - Page 1083: 55

Answer

$${\text{ }}P\left( {9,2} \right) = \frac{1}{{2520}}$$

Work Step by Step

$$\eqalign{ & P\left( {9,2} \right) \cr & {\text{Use the permutation formula }}P\left( {n,r} \right) = \frac{{n!}}{{\left( {n - r} \right)!}} \cr & {\text{ }}P\left( {9,2} \right) = \frac{{2!}}{{\left( {9 - 2} \right)!}} \cr & {\text{Evaluate and simplify}} \cr & {\text{ }}P\left( {9,2} \right) = \frac{{2!}}{{7!}} \cr & {\text{ }}P\left( {9,2} \right) = \frac{2}{{5040}} \cr & {\text{ }}P\left( {9,2} \right) = \frac{1}{{2520}} \cr} $$
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