#### Answer

5

#### Work Step by Step

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