Answer
The value of $\frac{n!}{\left( n-r \right)!\times r!}$ for $ n=8$ and $ r=3$ is $56$.
Work Step by Step
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 $56$.
