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: 57



Work Step by Step

Use the formula of permutation: $_{n}$P$_{r}$=$\frac{n!}{(n-r)!}$.You have two permutation problems:$_{4}$P$_{3}$ and $_{4}$P$_{2}$.Solve each one separately and divide the solutions in the end: -->$_{n}$P$_{r}$=$\frac{n!}{(n-r)!}$ $_{4}$P$_{3}$=$\frac{4!}{(4-3)!}$ -simplify- $_{4}$P$_{3}$=$\frac{4!}{1!}$ -write using factorial- $_{4}$P$_{3}$=$\frac{4*3*2*1}{1}$ -simplify- $_{4}$P$_{3}$=24 -->$_{n}$P$_{r}$=$\frac{n!}{(n-r)!}$ $_{4}$P$_{2}$=$\frac{4!}{(4-2)!}$ -simplify- $_{4}$P$_{2}$=$\frac{4!}{2!}$ -write using factorial- $_{4}$P$_{2}$=$\frac{4*3*2*1}{2*1}$ -simplify- $_{4}$P$_{2}$=12 Divide the solutions: 24$\div$12=2.So $_{4}$P$_{3}$ $\div$ $_{4}$P$_{2}$=2
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