Discrete Mathematics with Applications 4th Edition

Published by Cengage Learning
ISBN 10: 0-49539-132-8
ISBN 13: 978-0-49539-132-6

Chapter 5 - Sequences, Mathematical Induction, and Recursion - Exercise Set 5.1 - Page 244: 69

Answer

$n(n-1)...(n-k+1)\ or\ \prod_{i=n-k+1}^{n}(i)$

Work Step by Step

$\frac{n!}{(n-k)!}=\frac{n(n-1)...(n-k+1)(n-k)!}{(n-k)!}=n(n-1)...(n-k+1)=\prod_{i=n-k+1}^{n}(i)$
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