Answer
10
Work Step by Step
Use the formula of combination: $_{n}$C$_{r}$=$\frac{n!}{r!(n-r)!}$. Plug in 5 for N and 3 for R:
$_{n}$C$_{r}$=$\frac{n!}{r!(n-r)!}$
$_{5}$C$_{3}$=$\frac{5!}{3!(5-3)!}$ -simplify like terms-
$_{5}$C$_{3}$=$\frac{5!}{3! (2!)}$ -write using factorial-
$_{5}$C$_{3}$=$\frac{5*4*3*2*1}{(3*2*1)(2*1)}$ -simplify-
$_{5}$C$_{3}$=10
