Answer
$4$
Work Step by Step
Use the formula $C(n,r)=\dfrac{n!}{r!(n-r)!}$ to obtain:
$C(4,3)=\dfrac{4!}{3!(4-3)!}$
$C(4,3)=\dfrac{4!}{3!(1!)}$
$C(4,3)=\dfrac{ 4 {\times} 3 {\times} 2 {\times} 1}{(3{\times}2 {\times} 1)(1)}$
Simplify to get
$C(4,3)=4$
There are $4$ possible combinations.