## Algebra 2 (1st Edition)

$30$
We know that if we want to select $r$ objects out of $n$ disregarding the order, we can do this in $_nC_r=\frac{n!}{r!(n-r)!}$ ways. This gives: $_{5}C_2\cdot_{3}C_1=\frac{5!}{3!2!}\frac{3!}{2!1!}=10\cdot3=30$