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

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

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.

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