## Precalculus (6th Edition) Blitzer

The first three finishers come in from a competition of 6 automobiles. Te order in which the finishers win the race matters because each finisher will finish with a different rank and there are no ties. Therefore, we need to find the number of permutations of 6 things taken 3 at a time. Apply the formula, ${}_{n}{{P}_{r}}=\frac{n!}{\left( n-r \right)!}$ Where $n=6,r=3$ \begin{align} & {}_{6}{{P}_{3}}=\frac{6!}{\left( 6-3 \right)!} \\ & =\frac{6!}{3!} \end{align} Solve further the solution to get, \begin{align} & _{6}{{P}_{3}}=\frac{6\times 5\times 4\times 3!}{3!} \\ & =6\times 5\times 4 \\ & =120 \end{align}