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: 22


The required solution is $\frac{{}_{20}{{P}_{2}}}{2!}-{}_{20}{{C}_{2}}=0$.

Work Step by Step

The permutation and combination formulas respectively are: ${}_{n}{{P}_{r}}=\frac{n!}{\left( n-r \right)!}$ And ${}_{n}{{C}_{r}}=\frac{n!}{r!\left( n-r \right)!}$ Applying these formulas into the provided expression, we get: $\begin{align} & \frac{{}_{20}{{P}_{2}}}{2!}-{}_{20}{{C}_{2}}=\frac{20!}{2!\left( 20-2 \right)!}-\frac{20!}{2!\left( 20-2 \right)!} \\ & =0 \end{align}$ $\frac{{}_{20}{{P}_{2}}}{2!}-{}_{20}{{C}_{2}}=0$
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