Chapter 12.6, Problem 22E

To determine

To find: Evaluation of $\left(\begin{array}{l}5\\ 2\end{array}\right)\left(\begin{array}{c}5\\ 3\end{array}\right)$ using pascal’s triangle.

Answer to Problem 22E

  $\Rightarrow \left(\begin{array}{l}5\\ 2\end{array}\right)\left(\begin{array}{l}5\\ 3\end{array}\right)=10\times 10=100$

Explanation of Solution

Given information:

  $\left(\begin{array}{l}5\\ 2\end{array}\right)\left(\begin{array}{c}5\\ 3\end{array}\right)$

Calculation:

The formula for the expression $\left(\begin{array}{l}n\\ k\end{array}\right)$ is,

  $\left(\begin{array}{l}n\\ k\end{array}\right)=\frac{n!}{\left(n-k\right)!k!}$

  $\begin{array}{l}\Rightarrow \left(\begin{array}{l}5\\ 2\end{array}\right)=\frac{5!}{\left(5-2\right)!2!}\\ \Rightarrow \left(\begin{array}{l}5\\ 2\end{array}\right)=\frac{5\times 4\times 3\times 2\times 1}{\left(3\times 2\times 1\right)\left(2\times 1\right)}\\ \Rightarrow \left(\begin{array}{l}5\\ 2\end{array}\right)=\frac{5\times 4}{\left(2\times 1\right)}\\ \Rightarrow \left(\begin{array}{l}5\\ 2\end{array}\right)=10\\ \Rightarrow \left(\begin{array}{l}5\\ 3\end{array}\right)=\frac{5!}{\left(5-3\right)!3!}\\ \Rightarrow \left(\begin{array}{l}5\\ 3\end{array}\right)=\frac{5\times 4\times 3\times 2\times 1}{\left(3\times 2\times 1\right)\left(2\times 1\right)}\\ \Rightarrow \left(\begin{array}{l}5\\ 3\end{array}\right)=\frac{5\times 4}{\left(2\times 1\right)}\\ \Rightarrow \left(\begin{array}{l}5\\ 3\end{array}\right)=10\end{array}$

  $\Rightarrow \left(\begin{array}{l}5\\ 2\end{array}\right)\left(\begin{array}{l}5\\ 3\end{array}\right)=10\times 10=100$