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A group consists of 4 girls and 7 boys.In how many ways can a team of 5 members be selected if the team has no girls

$\begin{array}{1 1}(A)\;20\\(B)\;21\\(C)\;22\\(D)\;23\end{array} $

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  • $C(n,r)=\large\frac{n!}{r!(n-r)!}$
Total no of girls =4
Total no of boys =7
Total members in the team =5
$\therefore$ Total no of ways of selecting the team with 5 members which does not include girls =$7C_5$
$7C_5=\large\frac{7!}{5!\times 2!}$
$\Rightarrow \large\frac{7\times 6\times 5!}{2\times 1\times 5!}$
$\Rightarrow 21$ ways
Hence (B) is the correct answer.
answered May 15, 2014 by sreemathi.v

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