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: 12

Answer

The required solution is $792$

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 $_{12}{{C}_{5}}$. Here, $ n=12,r=5$. Put the value of n, r in the above formula. Then: $\begin{align} & _{12}{{C}_{5}}=\frac{12!}{\left( 12-5 \right)!5!} \\ & =\frac{12!}{7!5!} \\ & =\frac{12\cdot 11\cdot 10\cdot 9\cdot 8\cdot 7!}{7!\left( 5\cdot 4\cdot 3\cdot 2\cdot 1 \right)} \end{align}$ And simplify further, $\begin{align} & \frac{12\cdot 11\cdot 10\cdot 9\cdot 8\cdot 7!}{7!\left( 5\cdot 4\cdot 3\cdot 2\cdot 1 \right)}=\frac{12\cdot 11\cdot 10\cdot 9\cdot 8}{5\cdot 4\cdot 3\cdot 2\cdot 1} \\ & =\frac{95040}{120} \\ & =792 \end{align}$ Hence, $_{12}{{C}_{5}}=792$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.