College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 8 - Sequences, Induction, and Probability - Exercise Set 8.6 - Page 770: 58

Answer

The number of selections $40_{{C}{_{8}}}$ = 76,904,685

Work Step by Step

Total number of books in the list = 40 Number of books have to choose = 8 To find the number of ways of selection we determine by the formula $n_{{C}{_{r}}}$ = $\frac{n!}{(n - r)! r!}$ The number of selections $40_{{C}{_{8}}}$ = $\frac{40!}{(40 - 8)! 8!}$ = $\frac{40!}{32! . 8!}$ = $\frac{40\times39\times38\times37\times36\times35\times34\times33\times32!}{32! .(8\times7\times6\times5\times4\times3\times2\times1)}$ = $\frac{40\times39\times38\times37\times36\times35\times34\times33}{8\times7\times6\times5\times4\times3\times2\times1}$ = 76904685
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.