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: 57



Work Step by Step

Use the formula of permutation:.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)!}$ $_{4}$$P_{3}$=$\frac{4!}{1!}$ $_{4}$$P_{3}$=$\frac{4*3*2*1}{1}$ $_{4}$$P_{3}$=24 --> $_{n}$$P_{r}$=$\frac{n!}{(n-r)!}$ $_{4}$$P_{2}$=$\frac{4!}{(4-2)!}$ $_{4}$$P_{2}$=$\frac{4!}{2!}$ $_{4}$$P_{2}$=$\frac{4*3*2*1}{2*1}$ $_{4}$$P_{2}$=12 Divide the solutions: 24÷12=2.So $_{4}$$P_{3}$ $\div$ $_{4}$$P_{2}$=2
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