$\begin{array}{1 1}(A)\;1550\\(B)\;1440\\(C)\;1444\\(D)\;1500\end{array} $

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

First women can take any of the chairs marked 1 to 4 in 4 different way.

Second women can take any of the remaining 3 chairs from those marked 1 to 4 in 3 different ways.

So,total no of ways in which women can take seat =$4\times 3$

$\Rightarrow 4P_2$

$4P_2=\large\frac{4!}{(4-2)!}$

$\qquad=\large\frac{4\times 3\times 2\times 1}{2\times 1}$

$\qquad =12$

After two women are seated 6 chairs remains

First man take seat in any of the 6 chairs in 6 different ways,second man can take seat in any of the remaining 5 chairs in 5 different ways

Third man can take seat in any of the remaining 4 chairs in 4 different ways.

So,total no of ways in which men can take seat =$6\times 5\times 4$

$\Rightarrow 6P_3$

$6P_3=\large\frac{6!}{(6-3)!}$

$\Rightarrow \large\frac{6\times 5 \times 4\times 3\times 2\times 1}{3\times 2\times 1}$

$\Rightarrow 120$

Hence total number of ways in which men and women can be seated =$120\times 12$

$\Rightarrow 1440$

Hence (B) is the correct answer.

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