Answer
$15$
Work Step by Step
Use the formula $C(n,r)=\dfrac{n!}{r!(n-r)!}$ to obtain:
$3 (_5C_4)=3\left(\dfrac{5!}{5!(5-4)!}\right)$
$3(_5C_4)=3\left(\dfrac{5!}{5!1!}\right)$
$3(_5C_4)=3\left(\dfrac{5{\times}4 {\times} 3 {\times} 2 {\times} 1}{(4 {\times}3{\times}2 {\times} 1)(1)}\right)$
$\require{cancel}
3(_5C_4)=3\left(\dfrac{5{\times}\cancel{4 {\times} 3 {\times} 2 {\times} 1}}{(\cancel{4 {\times}3{\times}2 {\times} 1})(1)}\right)$
Simplify to get :
$3(_{5}C_{4})=3(5)=15$
There are $15$ possible combinations.