Answer
See below.
Work Step by Step
We know that ${n\choose j}=\frac{n!}{(n-j)!j!}$ by definition.
Hence here plugging in $n=8,j=2$: ${8\choose 2}=\frac{8!}{(8-2)!2!}=\frac{8!}{6!2!}$
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 - Concept and Vocabulary Check - Page 757: 2
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