Precalculus (10th Edition)

Published by Pearson

Chapter 13 - Counting and Probability - 13.2 Permutations and Combinations - 13.2 Asses Your Understanding - Page 856: 50

Answer

$5209344$

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 $C(4,2)C(8,3)C(20,5)=\frac{4\cdot3}{2\cdot1}\frac{8\cdot7\cdot6}{3\cdot2\cdot1}\frac{20\cdot19\cdot18\cdot17\cdot16}{5\cdot4\cdot3\cdot2\cdot1}=6\cdot56\cdot15504=5209344$.

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.