## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 10 - Section 10.6 - Counting Principles, Permutations, and Combinations - Exercise Set - Page 1103: 11

#### Answer

The required solution is $330$

#### 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 $_{11}{{C}_{4}}$. Here, $n=11,r=4$. Put the value of n, r in the above formula. Then: \begin{align} & _{11}{{C}_{4}}=\frac{11!}{\left( 11-4 \right)!4!} \\ & =\frac{11!}{7!4!} \\ & =\frac{11\cdot 10\cdot 9\cdot 8\cdot 7!}{7!\cdot 4\cdot 3\cdot 2\cdot 1} \end{align} Simplifying further, \begin{align} & \frac{11\cdot 10\cdot 9\cdot 8\cdot 7!}{7!\cdot 4\cdot 3\cdot 2\cdot 1}=\frac{11\cdot 10\cdot 9\cdot 8}{4\cdot 3\cdot 2\cdot 1} \\ & =\frac{7920}{24} \\ & =330 \end{align} Hence, $_{11}{{C}_{4}}=330$

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