College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 8 - Sequences, Induction, and Probability - Exercise Set 8.5 - Page 759: 83

Answer

See below.

Work Step by Step

We know that ${n\choose r}=\frac{n!}{(n-r)!r!}$ by definition Hence here: ${n\choose r}=\frac{n!}{(n-r)!r!}=\frac{n!}{r!(n-r)!}=\frac{n!}{(n-(n-r))!(n-r)!}={n\choose n-r}$. Thus we proved what we wanted to.
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