Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 13 - Counting and Probability - 13.2 Permutations and Combinations - 13.2 Asses Your Understanding - Page 855: 42



Work Step by Step

The order of the professors doesn't matter here, so the situation involves combination. We know that $C(n,r)=\dfrac{n!}{(n-r)!r!}$. Thus, with $n=8$ and $r=3$, then $C(8,3)=\dfrac{8!}{(8-3)!\cdot3!}=\dfrac{8\cdot7\cdot6\cdot5!}{5!\cdot3\cdot2\cdot1}=\dfrac{8\cdot7\cdot6}{6}=56$.
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.