## Precalculus (6th Edition) Blitzer

The value of $\frac{n!}{\left( n-r \right)!\times r!}$ for $n=8$ and $r=3$ is $56$.
The given expression is $\frac{n!}{\left( n-r \right)!\times r!}$ Put $n=8$ and $r=3$ in the expression $\frac{n!}{\left( n-r \right)!\times r!}$ and simplify, \begin{align} & \frac{n!}{\left( n-r \right)!\times r!}=\frac{8!}{\left( 8-3 \right)!\times 3!} \\ & =\frac{8!}{5!\times 3!} \end{align} It is known that \begin{align} & n!=n\times \left( n-1 \right)\times \left( n-2 \right)\times \left( n-3 \right)..........3\times 2\times 1 \\ & =n\times \left( n-1 \right)! \end{align} \begin{align} & n!=\frac{8\times 7\times 6\times 5!}{5!\times 3\times 2\times 1} \\ & =\frac{8\times 7\times 6}{3\times 2\times 1} \\ & =56 \end{align} Thus, the value of $\frac{n!}{\left( n-r \right)!\times r!}$ for $n=8$ and $r=3$ is $56$.