Algebra 1

Published by Prentice Hall
ISBN 10: 0133500403
ISBN 13: 978-0-13350-040-0

Chapter 12 - Data Analysis and Probability - 12-6 Permutations and Combinations - Practice and Problem-Solving Exercises - Page 756: 58

Answer

5

Work Step by Step

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