## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 10 - Section 10.6 - Counting Principles, Permutations, and Combinations - Exercise Set - Page 1103: 26

#### Answer

The required solution is $\frac{{}_{10}{{C}_{3}}}{{}_{6}{{C}_{4}}}-\frac{46!}{44!}=-2062$

#### 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: $\frac{{}_{10}{{C}_{3}}}{{}_{6}{{C}_{4}}}-\frac{46!}{44!}=\frac{\frac{10!}{3!\left( 10-3 \right)!}}{\frac{6!}{4!\left( 6-4 \right)!}}-\frac{46!}{44!}$ $=\frac{10!}{3!\times 7!}\times \frac{4!\times 2!}{6!}-\frac{46!}{44!}$ $=\frac{10\times 9\times 8\times 7!}{7!\times 3\times 2}\times \frac{4!\times 2\times 1}{6\times 5\times 4!}-\frac{46\times 45\times 44!}{44!}$ \begin{align} & =8-2070 \\ & =2062 \\ \end{align} $\frac{{}_{10}{{C}_{3}}}{{}_{6}{{C}_{4}}}-\frac{46!}{44!}=-2062$

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