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