Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 10 - Section 10.6 - Counting Principles, Permutations, and Combinations - Exercise Set - Page 1103: 9

Answer

The required solution is $126$

Work Step by Step

We know that the representation $_{n}{{C}_{r}}$ counts the number of assortments of n items taken r at a time. And the number of assortment of n items taken r at a time can be evaluated as: $_{n}{{C}_{r}}=\frac{n!}{\left( n-r \right)!r!}$ And the provided expression is $_{9}{{C}_{5}}$. Here, $ n=9,r=5$. Put the value of n, r in the above formula. Then: $\begin{align} & _{9}{{C}_{5}}=\frac{9!}{\left( 9-5 \right)!5!} \\ & =\frac{9!}{4!5!} \\ & =\frac{9\cdot 8\cdot 7\cdot 6\cdot 5!}{4\cdot 3\cdot 2\cdot 1\cdot 5!} \end{align}$ And simplify further, $\begin{align} & \frac{9\cdot 8\cdot 7\cdot 6\cdot 5!}{4\cdot 3\cdot 2\cdot 1\cdot 5!}=\frac{9\cdot 8\cdot 7\cdot 6}{4\cdot 3\cdot 2\cdot 1} \\ & =\frac{3024}{24} \\ & =126 \end{align}$ Hence, $_{9}{{C}_{5}}=126$
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