Answer
$n=12$
Work Step by Step
$_{n}P_k=\frac{n!}{(n-k)!}$
Hence here $_{n}P_6=\frac{n!}{(n-6)!}=12_{n-1}P_5=12\frac{(n-1)!}{((n-1)-5)!}\\\frac{n!}{(n-6)!}=12\frac{(n-1)!}{(n-6)!}\\n(n-1)(n-2)(n-3)(n-4)(n-5)=12(n-1)(n-2)(n-3)(n-4)(n-5)\\n=12$
Algebra and Trigonometry 10th Edition
by Larson, Ron
Published by
Cengage Learning
ISBN 10:
9781337271172
ISBN 13:
978-1-33727-117-2
Chapter 11 - 11.6 - Counting Principles - 11.6 Exercises - Page 825: 82
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