College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 8, Sequences and Series - Section 8.6 - The Binomial Theorem - 8.6 Exercises: 21

Answer

$\displaystyle \left(\begin{array}{l} 3\\ 1 \end{array}\right)\left(\begin{array}{l} 4\\ 2 \end{array}\right)=18$

Work Step by Step

We evaluate: $\displaystyle \left(\begin{array}{l} 3\\ 1 \end{array}\right)\left(\begin{array}{l} 4\\ 2 \end{array}\right)=\frac{3!}{1!2!}\frac{4!}{2!2!}=\frac{3*2!*4*3*2!}{1*2!*2!*2*1}=\frac{3*3*4}{2}=\frac{9*4}{2}=18$
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