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 856: 59

Answer

$660$

Work Step by Step

If we have two groups with $n$ and $m$ objects and we choose $r(r\leq n)$ and $s(s\leq m)$ objects from the two groups where the order doesn't matter then the number of combinations is: $C(n,r)C(m,s).$ (Similarly for more than $2$ groups.) We know that $C(n,r)=\frac{n(n-1)(n-2)...(n-k+1)}{r!}$. Also $C(n,0)=1$ by convention. Also $C(n,r)=C(n,n-r).$ Hence here because out of the $4$ pitchers $1$ is chosen and from the rest $11$, $9-1=8$ are chosen$C(4,1)C(11,8)=\frac{4}{1}C(11,3)=4\cdot\frac{11\cdot10\cdot9}{3\cdot2\cdot1}=4\cdot165=660$
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.