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: 25


The required solution is $\frac{{}_{7}{{C}_{3}}}{{}_{5}{{C}_{4}}}-\frac{98!}{96!}=-9499$

Work Step by Step

The combination formula is: ${}_{n}{{C}_{r}}=\frac{n!}{r!\left( n-r \right)!}$ Applying this formulas into the provided expression, we get: $\frac{{}_{7}{{C}_{3}}}{{}_{5}{{C}_{4}}}-\frac{98!}{96!}=\frac{\frac{7!}{3!\left( 7-3 \right)!}}{\frac{5!}{4!\left( 5-4 \right)!}}-\frac{98!}{96!}$ $=\frac{7!}{3!\times 4!}\times \frac{4!\times 1!}{5!}-\frac{98!}{96!}$ $=\frac{7\times 6}{3\times 2}-\frac{98\times 97\times 96!}{96!}$ $\begin{align} & =7-9506 \\ & =-9499 \\ \end{align}$ $\frac{{}_{7}{{C}_{3}}}{{}_{5}{{C}_{4}}}-\frac{98!}{96!}=-9499$
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