Answer
1365
Work Step by Step
Use the formula of combination: $_{n}$C$_{r}$=$\frac{n!}{r!(n-r)!}$. Plug in 15 for N and 4 for R:
$_{n}$C$_{r}$=$\frac{n!}{r!(n-r)!}$
$_{15}$C$_{4}$=$\frac{15!}{4!(15-4)!}$ -simplify like terms-
$_{15}$C$_{4}$=$\frac{15!}{4! (11!)}$ -write using factorial-
$_{15}$C$_{4}$=$\frac{15*14*13*12*11*10*9*8*7*6*5*4*3*2*1}{(4*3*2*1)(11*10*9*8*7*6*5*4*3*2*1)}$ -simplify-
$_{15}$C$_{4}$=1365
