Answer
$132$
Work Step by Step
Since the ordering matters in the selection, we use permutations.
$_nP_r=\frac{n!}{(n-r)!}$, hence here: $_{12}P_2=\frac{12!}{(12-2)!}=\frac{12!}{10!}=12\cdot11=132$
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 Binomial Theorem - 10.2 Exercises - Skill Practice - Page 695: 38
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