Algebra 1: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281140
ISBN 13: 978-0-13328-114-9

Chapter 12 - Data Analysis and Probability - 12-6 Permutations and Combinations - Practice and Problem-Solving Exercises - Page 768: 58



Work Step by Step

Use the formula of permutation: $_{n}$P$_{r}$=$\frac{n!}{(n-r)!}$.You have two permutation problems:$_{7}$P$_{3}$ and $_{7}$P$_{2}$.Solve each one seperately and divide the solutions in the end: -->$_{n}$P$_{r}$=$\frac{n!}{(n-r)!}$ $_{7}$P$_{3}$=$\frac{7!}{(7-3)!}$ -simplify- $_{7}$P$_{3}$=$\frac{7!}{4!}$ -write using factorial- $_{7}$P$_{3}$=$\frac{7*6*5*4*3*2*1}{4*3*2*1}$ -simplify- $_{7}$P$_{3}$=210 -->$_{n}$P$_{r}$=$\frac{n!}{(n-r)!}$ $_{7}$P$_{2}$=$\frac{7!}{(7-2)!}$ -simplify- $_{7}$P$_{2}$=$\frac{7!}{5!}$ -write using factorial- $_{7}$P$_{2}$=$\frac{7*6*5*4*3*2*1}{5*4*3*2*1}$ -simplify- $_{7}$P$_{2}$=42 Divide the solutions: 210$\div$42=5.So $_{7}$P$_{3}$ $\div$ $_{7}$P$_{2}$=5
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