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 1104: 65


The number of cones that can be formed with three different flavors from 15 total flavors is $2730$.

Work Step by Step

We know that: ${}_{n}{{P}_{r}}=\frac{n!}{\left( n-r \right)!}$ So, from the information, $\begin{align} & n=15 \\ & r=3 \end{align}$ Then, we have to find the number of permutations of 15 things taken 3 at a time: $\begin{align} & {}_{15}{{P}_{3}}=\frac{15!}{\left( 15-3 \right)!} \\ & =\frac{15!}{12!} \\ & =\frac{15\times 14\times 13\times 12!}{12!} \\ & =2,730 \end{align}$
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