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: 28


The required solution is $\frac{{}_{5}{{C}_{1}}.{}_{7}{{C}_{2}}}{{}_{12}{{C}_{3}}}=\frac{21}{44}$ .

Work Step by Step

The combination formula ${}_{n}{{C}_{r}}=\frac{n!}{r!\left( n-r \right)!}$. Applying this formulas into the provided expression, we get: $\frac{{}_{5}{{C}_{1}}.{}_{7}{{C}_{2}}}{{}_{12}{{C}_{3}}}=\frac{\frac{5!}{1!\left( 5-1 \right)!}\times \frac{7!}{2!\left( 7-2 \right)!}}{\frac{12!}{3!\left( 12-3 \right)!}}$ $=\frac{5!\times 7!}{4!\times 2!\times 5!}\times \frac{3!\times 9!}{12!}$ $=\frac{7\times 6\times 5\times 4!}{4!\times 2}\times \frac{9!\times 3\times 2\times 1}{12\times 11\times 10\times 9!}$ $=\frac{21}{44}$ $\frac{{}_{5}{{C}_{1}}.{}_{7}{{C}_{2}}}{{}_{12}{{C}_{3}}}=\frac{21}{44}$
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