One pair one different arrangement can be$\rightarrow 1C_1\times 4C_1\times \large\frac{3!}{2!}$
A
$\Rightarrow 1\times 4\times 3=12$
Three different $\rightarrow 5C_3$
Total arrangements $\rightarrow 5C_3\times 3!$
$\Rightarrow \large\frac{5!}{3!2!}$$\times 3!$
$\Rightarrow \large\frac{5\times 4}{2}$$\times 6$
$\Rightarrow 60$
$\therefore$ The required no of ways =72
Hence (A) is the correct answer.