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


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.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.