Chapter 8, Sequences and Series - Section 8.6 - The Binomial Theorem - 8.6 Exercises - Page 636: 21

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

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