Answer
2
Work Step by Step
Use the formula of permutation:.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)!}$
$_{4}$$P_{3}$=$\frac{4!}{1!}$
$_{4}$$P_{3}$=$\frac{4*3*2*1}{1}$
$_{4}$$P_{3}$=24
--> $_{n}$$P_{r}$=$\frac{n!}{(n-r)!}$
$_{4}$$P_{2}$=$\frac{4!}{(4-2)!}$
$_{4}$$P_{2}$=$\frac{4!}{2!}$
$_{4}$$P_{2}$=$\frac{4*3*2*1}{2*1}$
$_{4}$$P_{2}$=12
Divide the solutions: 24÷12=2.So $_{4}$$P_{3}$ $\div$ $_{4}$$P_{2}$=2
