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The number of ways in which we can choose a committee from four men and six women so that the committee includes at least two men and exactly twice as many women and men is

$\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$
$4C_2\times 6C_4+4C_3\times 6C_6=6\times 15+4\times 1$
$\Rightarrow 90+4$
$\Rightarrow 94$
Hence (A) is the correct answer.
answered May 16, 2014 by sreemathi.v

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