Answer
56,combination
Work Step by Step
Use the formula for combination: $_{n}$C$_{r}$=$\frac{n!}{r!(n-r)!}$ because arrangement or order do not matter.Plug in 8 for N and 3 for R:
$_{n}$C$_{r}$=$\frac{n!}{r!(n-r)!}$
$_{8}$C$_{3}$=$\frac{8!}{8!(8-3)!}$ -simplify like terms-
$_{8}$C$_{3}$=$\frac{8!}{3! (5!)}$ -write using factorial-
$_{8}$C$_{3}$=$\frac{8*7*6*5*4*3*2*1}{(3*2*1)(5*4*3*2*1)}$ -simplify-
$_{8}$C$_{3}$=56
