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 8.6 - Page 771: 93



Work Step by Step

If we want to choose $k$ elements out of $n$ disregarding the order, not allowing repetition, we can do this in $_{n}C_k=\frac{n!}{(n-k)!k!}$ ways. The order doesn't matter here in the questions, thus we have to use combinations. Thus $_{10}C_{8}\cdot_{5}C_{3}=\frac{10!}{(10-8)!8!}\frac{5!}{(5-3)!3!}=45\cdot10=450$
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