Answer
See below.
Work Step by Step
We know that ${n\choose j}=\frac{n!}{(n-j)!j!}$.
Hence here ${5\choose 3}=\frac{5!}{(5-3)!3!}=\frac{5\cdot4\cdot3\cdot2\cdot1}{2\cdot1\cdot3\cdot2\cdot1}=10$.
College Algebra (6th Edition)
by Blitzer, Robert F.
Published by
Pearson
ISBN 10:
0-32178-228-3
ISBN 13:
978-0-32178-228-1
Chapter 8 - Sequences, Induction, and Probability - Exercise Set 8.5 - Page 758: 59
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