Answer
3
Work Step by Step
Use the formula of permutation: $_{n}$P$_{r}$=$\frac{n!}{(n-r)!}$.You have two permutation problems:$_{5}$P$_{3}$ and $_{5}$P$_{2}$.Solve each one seperately and divide the solutions in the end:
-->$_{n}$P$_{r}$=$\frac{n!}{(n-r)!}$
$_{5}$P$_{3}$=$\frac{5!}{(5-3)!}$ -simplify-
$_{5}$P$_{3}$=$\frac{5!}{2!}$ -write using factorial-
$_{5}$P$_{3}$=$\frac{5*4*3*2*1}{2*1}$ -simplify-
$_{5}$P$_{3}$=60
-->$_{n}$P$_{r}$=$\frac{n!}{(n-r)!}$
$_{5}$P$_{2}$=$\frac{5!}{(5-2)!}$ -simplify-
$_{5}$P$_{2}$=$\frac{5!}{3!}$ -write using factorial-
$_{5}$P$_{2}$=$\frac{5*4*3*2*1}{3*2*1}$ -simplify-
$_{5}$P$_{2}$=20
Divide the solutions: 60$\div$20=3.So $_{5}$P$_{3}$ $\div$ $_{5}$P$_{2}$=3
