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: 14

Answer

The required solution is $1$.

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 $_{4}{{C}_{4}}$. Here, $ n=4,r=4$. Put the value of n, r in the above formula. Then: $\begin{align} & _{4}{{C}_{4}}=\frac{4!}{\left( 4-4 \right)!4!} \\ & =\frac{4!}{0!4!} \\ & =\frac{1}{0!} \end{align}$ Since, $0!=1$ So, $\begin{align} & \frac{1}{0!}=\frac{1}{1} \\ & =1 \end{align}$ Hence, $_{4}{{C}_{4}}=1$
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