## Precalculus (6th Edition) Blitzer

The value of $\frac{n!}{\left( n-r \right)!}$ for $n=20$ and $r=3$ is $6840$.
The given expression is $\frac{n!}{\left( n-r \right)!}$. Put $n=20$ and $r=3$ in the expression $\frac{n!}{\left( n-r \right)!}$ and simplify, \begin{align} & \frac{n!}{\left( n-r \right)!}=\frac{20!}{\left( 20-3 \right)!} \\ & =\frac{20!}{17!} \end{align} It is know 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{20\times 19\times 18\times 17!}{17!} \\ & =20\times 19\times 18 \\ & =6,840 \end{align} Thus, the value of $\frac{n!}{\left( n-r \right)!}$ for $n=20$ and $r=3$ is $6840$.