$\begin{array}{1 1}(A)\;94\\(B)\;126\\(C)\;128\\(D)\;\text{None}\end{array} $

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

Total no of men =4

Total no of women =6

The total no of ways =$4C_2\times 6C_4+4C_3\times 6C_6$

$4C_2=\large\frac{4!}{2!2!}=\frac{4\times 3\times 2!}{2\times 2!}$$=6$

$6C_4=\large\frac{6!}{4!2!}=\frac{6\times 5\times 4!}{4!\times 2}$$=15$

$4C_3=\large\frac{4!}{3!1!}=\frac{4\times 3!}{3!\times 1}$$=4$

$6C_6=\large\frac{6!}{6!0!}$$=1$

$4C_2\times 6C_4+4C_3\times 6C_6=6\times 15+4\times 1$

$\Rightarrow 90+4$

$\Rightarrow 94$

Hence (A) is the correct answer.

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