Answer
5
Work Step by Step
Use the formula of permutation: $_{n}$P$_{r}$=$\frac{n!}{(n-r)!}$.You have two permutation problems:$_{7}$P$_{3}$ and $_{7}$P$_{2}$.Solve each one seperately and divide the solutions in the end:
-->$_{n}$P$_{r}$=$\frac{n!}{(n-r)!}$
$_{7}$P$_{3}$=$\frac{7!}{(7-3)!}$ -simplify-
$_{7}$P$_{3}$=$\frac{7!}{4!}$ -write using factorial-
$_{7}$P$_{3}$=$\frac{7*6*5*4*3*2*1}{4*3*2*1}$ -simplify-
$_{7}$P$_{3}$=210
-->$_{n}$P$_{r}$=$\frac{n!}{(n-r)!}$
$_{7}$P$_{2}$=$\frac{7!}{(7-2)!}$ -simplify-
$_{7}$P$_{2}$=$\frac{7!}{5!}$ -write using factorial-
$_{7}$P$_{2}$=$\frac{7*6*5*4*3*2*1}{5*4*3*2*1}$ -simplify-
$_{7}$P$_{2}$=42
Divide the solutions: 210$\div$42=5.So $_{7}$P$_{3}$ $\div$ $_{7}$P$_{2}$=5
