Thinking Mathematically (6th Edition)

Published by Pearson
ISBN 10: 0321867327
ISBN 13: 978-0-32186-732-2

Chapter 11 - Counting Methods and Probability Theory - 11.2 Permutations - Exercise Set 11.2: 27

Answer

6

Work Step by Step

See Factorial Notation, page 696: $n!=n(n-1)(n-2)\cdots(3)(2)(1)$ and, by definition, $0!=1$ ------------ $(\displaystyle \frac{12}{4})!=(3)!=3\times 2\times 1=6$
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