Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 11 - Section 11.5 - The Binomial Theorem - Exercise Set - Page 664: 50

Answer

$4$

Work Step by Step

Since $\left(\begin{array}{c} n\\ r\end{array} \right)=\dfrac{n!}{r!(n-r)!}$, then, $\left(\begin{array}{c} 4 \\ 3 \end{array} \right)$ is equal to \begin{array}{l}\require{cancel} \dfrac{4!}{3!(4-3)!} \\\\= \dfrac{4!}{3!(1)!} \\\\= 4 \end{array}
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