Answer
$132$
Work Step by Step
We know that $n!=n\cdot(n-1)\cdot(n-2)...\cdot2\cdot1.$
Hence,
$\require{cancel}
\dfrac{12!}{10!}\\
=\dfrac{12\cdot11\cdot10!}{10!}\\
=\dfrac{12\cdot11\cdot\cancel{10!}}{\cancel{10!}}
\\=12\cdot11\\
=132$
Precalculus (10th Edition)
by Sullivan, Michael
Published by
Pearson
ISBN 10:
0-32197-907-9
ISBN 13:
978-0-32197-907-0
Chapter 12 - Sequences; Induction; the Binomial Theorem - 12.1 Sequences - 12.1 Assess Your Understanding - Page 807: 12
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