$\begin{array}{1 1}(A)\;72\\(B)\;120\\(C)\;14\\(D)\;\text{None of these}\end{array} $

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- $C(n,r)=\large\frac{n!}{r!(n-r)!}$

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.

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