$\begin{array}{1 1}(A)\;90\\(B)\;91\\(C)\;92\\(D)\;93\end{array} $

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

Since the team has to consist of at least three girls

The team can consist of

(a) 3 girls and 2 boys

(b) 4 girls and 1 boy

Note : The team cannot have all 5 girls,because the group has only 4 girls.

3 girls and 2 boys can be selected in =$4C_3\times 7C_2$ ways

4 girls and 1 boy can be selected in =$4C_4\times 7C_1$ ways

$\therefore$ The required number of ways =$4C_3\times 7C_2+4C_4\times 7C_1$

$4C_3=\large\frac{4!}{3!1!}=\frac{4\times 3!}{3!}$$=4$

$7C_2=\large\frac{7!}{2!5!}=\frac{7\times 6\times 5!}{2\times 5!}$$=21$

$4C_4=\large\frac{4!}{4!0!}$$=1$

$7C_1=\large\frac{7!}{1!6!}=\frac{7\times 6!}{6!}$$=7$

$\Rightarrow 4\times 21+1\times 7$

$\Rightarrow 84+7$

$\Rightarrow 91$ ways

Hence (B) is the correct answer.

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