## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 10 - Section 10.6 - Counting Principles, Permutations, and Combinations - Exercise Set - Page 1103: 27

#### Answer

$\frac{{}_{4}{{C}_{2}}.{}_{6}{{C}_{1}}}{{}_{18}{{C}_{3}}}=\frac{3}{68}$.

#### 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: $\frac{{}_{4}{{C}_{2}}.{}_{6}{{C}_{1}}}{{}_{18}{{C}_{3}}}=\frac{\frac{4!}{2!\left( 4-2 \right)!}\times \frac{6!}{1!\left( 6-1 \right)!}}{\frac{18!}{3!\left( 18-3 \right)!}}$ $=\frac{4!\times 6!}{2!\times 2!\times 5!}\times \frac{3!\times 15!}{18!}$ $=\frac{4\times 3\times 2\times 6\times 5!}{5!\times 2\times 2}\times \frac{15!\times 3\times 2\times 1}{18\times 17\times 16\times 15!}$ $=\frac{3}{68}$ $\frac{{}_{4}{{C}_{2}}.{}_{6}{{C}_{1}}}{{}_{18}{{C}_{3}}}=\frac{3}{68}$

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