Answer
$15$
Work Step by Step
We know that $C(n,r)=\frac{n!}{(n-r)!r!}$and $C(n,0)=1$.
Hence,
$C(6,2)=\dfrac{6!}{(6-2)!\cdot2!}=\dfrac{6\cdot5\cdot4!}{4!\cdot2!}=\dfrac{6\cdot5}{2\cdot1}=15$
Precalculus (10th Edition)
by Sullivan, Michael
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 855: 18
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