$\begin{array}{1 1}(A)\;20\\(B)\;30\\(C)\;40\\(D)\;50\end{array} $

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

3 boys out of 5 boys can be selected in =$5C_3$ ways

3 girls out of 4 boys can be selected in =$4C_3$ ways

Number of boys in which 3 boys and 3 girls are selected are

$5C_3\times 4C_3=5C_2\times 4C_1$

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

$\Rightarrow \large\frac{20\times 4}{2}$$=40$

Hence (C) is the correct answer.

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