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In how many ways can a team of 3boys and 3 girls be selected from 5 boys and 4 girls?

$\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.
answered May 14, 2014 by sreemathi.v

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