Answer
$30$
Work Step by Step
We know that if we want to select $r$ objects out of $n$ disregarding the order, we can do this in $_nC_r=\frac{n!}{r!(n-r)!}$ ways.
This gives:
$_{5}C_2\cdot_{3}C_1=\frac{5!}{3!2!}\frac{3!}{2!1!}=10\cdot3=30$
Algebra 2 (1st Edition)
by Larson, Ron; Boswell, Laurie; Kanold, Timothy D.; Stiff, Lee
Published by
McDougal Littell
ISBN 10:
0618595414
ISBN 13:
978-0-61859-541-9
Chapter 10 Counting Methods and Probability - 10.2 Use Combinations and the Binomial Theorem - 10.2 Exercises - Problem Solving - Page 696: 48
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter 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.
