Answer
$91$
Work Step by Step
The order matters here, so it is basically $C(14,2).$
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 $C(14,2)=\frac{14\cdot13}{2!}=91$.
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 - Chapter Review - Review Exercises - Page 868: 16
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