Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 10 Counting Methods and Probability - 10.2 Use Combinations and Binomial Theorem - 10.2 Exercises - Skill Practice - Page 695: 44


See below.

Work Step by Step

We know that $_nC_r=\frac{n!}{r!(n-r)!}$. Hence $_nC_r\cdot_rC_m=\frac{n!}{r!(n-r)!}\frac{r!}{m!(r-m)!}=\frac{n!}{(n-r)!m!(r-m)!}$ $_nC_m\cdot_{n-m}C_{r-m}=\frac{n!}{m!(n-m)!}\frac{(n-m)!}{(r-m)!(n-m-(r-m))!}=\frac{n!}{m!(n-m)!}\frac{(n-m)!}{(r-m)!(n-r)!}=\frac{n!}{(n-r)!m!(r-m)!}$ Thus they are equal; thus we proved what we had to.
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