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: 23


The required solution is $1-\frac{{}_{3}{{P}_{2}}}{{}_{4}{{P}_{3}}}=\frac{3}{4}$ .

Work Step by Step

The permutation formula is: ${}_{n}{{P}_{r}}=\frac{n!}{\left( n-r \right)!}$ Applying these formulas into the provided expression, we get: $1-\frac{{}_{3}{{P}_{2}}}{{}_{4}{{P}_{3}}}=1-\frac{\frac{3!}{\left( 3-2 \right)!}}{\frac{4!}{\left( 4-3 \right)!}}=1-\frac{3!}{1!}\times \frac{1!}{4!}$ $=1-\frac{3\times 2\times 1}{4\times 3\times 2\times 1}$ $\begin{align} & =1-\frac{1}{4} \\ & =\frac{3}{4} \\ \end{align}$
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.