Answer
120
Work Step by Step
We know that:
$_nP_r = \frac{n!}{(n-r)!}$
This gives
$_6P_3 = \frac{6!}{(6-3)!}$
$\frac{6*5*4*3*2*1}{3*2*1}$
$6*5*4$
120
Algebra 1: Common Core (15th Edition)
by Charles, Randall I.
Published by
Prentice Hall
ISBN 10:
0133281140
ISBN 13:
978-0-13328-114-9
Chapter 12 - Data Analysis and Probability - 12-6 Permutations and Combinations - Lesson Check - Page 765: 3
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