Answer
2
Work Step by Step
Use the formula of permutation: $_{n}$P$_{r}$=$\frac{n!}{(n-r)!}$.You have two permutation problems:$_{4}$P$_{3}$ and $_{4}$P$_{2}$.Solve each one separately and divide the solutions in the end:
-->$_{n}$P$_{r}$=$\frac{n!}{(n-r)!}$
$_{4}$P$_{3}$=$\frac{4!}{(4-3)!}$ -simplify-
$_{4}$P$_{3}$=$\frac{4!}{1!}$ -write using factorial-
$_{4}$P$_{3}$=$\frac{4*3*2*1}{1}$ -simplify-
$_{4}$P$_{3}$=24
-->$_{n}$P$_{r}$=$\frac{n!}{(n-r)!}$
$_{4}$P$_{2}$=$\frac{4!}{(4-2)!}$ -simplify-
$_{4}$P$_{2}$=$\frac{4!}{2!}$ -write using factorial-
$_{4}$P$_{2}$=$\frac{4*3*2*1}{2*1}$ -simplify-
$_{4}$P$_{2}$=12
Divide the solutions: 24$\div$12=2.So $_{4}$P$_{3}$ $\div$ $_{4}$P$_{2}$=2
