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: 22

Answer

$\displaystyle \left(\begin{array}{l} 5\\ 2 \end{array}\right)\left(\begin{array}{l} 5\\ 3 \end{array}\right)=100$

Work Step by Step

We evaluate: $\displaystyle \left(\begin{array}{l} 5\\ 2 \end{array}\right)\left(\begin{array}{l} 5\\ 3 \end{array}\right)=\frac{5!}{2!3!}*\frac{5!}{3!2!}=\frac{5\cdot 4\cdot 3!}{2\cdot 1\cdot 3!}*\frac{5\cdot 4\cdot 3!}{3!\cdot 2\cdot 1}=\frac{5*4*5*4}{2*2}=5*2*5*2=10*10=100$
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