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# The number of parallelogram that can be formed from a set of four parallel lines intersecting another set of three parallel lines is

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

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.