Geometry: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281159
ISBN 13: 978-0-13328-115-6

Chapter 13 - Probability - 13-3 Permutations and Combinations - Practice and Problem-Solving Exercises - Page 841: 16

Answer

a) 1320 b) 95,040

Work Step by Step

a) We use the permutation equation: We know that: $_nP_r = \frac{n!}{(n-r)!}$ $_{12}P_3 = \frac{12!}{(12-3)!}$ $\frac{12!}{9!}$ We also know that: $x! = x(x-1)(x-2)...(1)$ Thus, we have: 1320 b) We know that: $_nP_r = \frac{n!}{(n-r)!}$ $_{12}P_5 = \frac{12!}{(12-5)!}$ $\frac{12!}{7!}$ We also know that: $x! = x(x-1)(x-2)...(1)$ Thus, we have: 95,040
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.