## Precalculus (6th Edition) Blitzer

The number of ways of forming a four-letter password is $840$.
We know that: ${}_{n}{{P}_{r}}=\frac{n!}{\left( n-r \right)!}$ So, from the information, \begin{align} & n=7 \\ & r=4 \\ \end{align} Then, we have to find the number of permutations of 7 things taken 4 at a time: \begin{align} & {}_{7}{{P}_{4}}=\frac{7!}{\left( 7-4 \right)!} \\ & =\frac{7!}{3!} \\ & =\frac{7\times 6\times \ldots \times 3!}{3!} \\ & =840 \end{align} Thus, the number of ways of forming a four-letter password is $840$.