$\begin{array}{1 1}(A)\;6\\(B)\;18\\(C)\;12\\(D)\;9\end{array} $

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Total no of parallel lines =4

Intersecting lines =3

$\therefore$ The required number of parallelogram =$4C_2\times 3C_2$

$4C_2=\large\frac{4!}{2!2!}=\frac{4\times 3\times 2!}{2\times 2!}$

$\Rightarrow 6$

$3C_2=\large\frac{3!}{2!1!}=\frac{3\times 2!}{2!}$

$\Rightarrow 3$

$4C_2\times 3C_2=6\times 3$

$\Rightarrow 18$

Hence (B) is the correct answer.

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