Answer
504
Work Step by Step
We know that:
$_nP_r = \frac{n!}{(n-r)!}$
$_9P_3 = \frac{9!}{(9-3)!}$
$\frac{9!}{6!}$
We also know that:
$x! = x(x-1)(x-2)...(1)$
Thus, we have:
$\frac{9*8*7*6*5*4*3*2*1}{6*5*4*3*2*1}$
= 504
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 - Practice and Problem-Solving Exercises - Page 766: 16
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