Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 11 - Further Topics in Algebra - 11.6 Basics of Counting Theory - 11.6 Exercises - Page 1061: 25

Answer

$495$

Work Step by Step

$C(n,r)=\displaystyle \frac{n!}{(n-r)!r!}$. Factorials: $n!=n(n-1)!=n(n-1)(n-2)!=...$ $n!=n(n-1)(n-2)\cdots(3)(2)(1)$ $ 0!=1,\qquad$ by definition. --- $C(12,4)=\displaystyle \frac{12!}{(12-4)!4!}=\frac{12\times 11\times 10\times 9\times(8!)}{(8!)\cdot 4!}$ $=\displaystyle \frac{[12]\times 11\times(10)\times 9}{[4\times 3]\times(2)\times 1}$ $=11\times 5\times 9$ = $495$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.