$\begin{array}{1 1}(A)\;430\\(B)\;440\\(C)\;441\\(D)\;550\end{array} $

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

Since at least one boy and one girl are to be there in every team

$\therefore$ The team consist of

(a) 1 boy 4 girls $\rightarrow 7C_1\times 4C_4$

(b) 2 boys and 3 girls $\rightarrow 7C_2\times 4C_3$

(c) 3 boys and 2 girls $\rightarrow 7C_3\times 4C_2$

(d) 4 boys and 1 girl $\rightarrow 7C_4\times 4C_1$

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

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

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

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

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

$7C_3=\large\frac{7!}{3!4!}=\frac{7\times 6\times 5\times 4!}{3\times 2\times 4!}$$=35$

$4C_2=\large\frac{4!}{2!2!}=\frac{4\times 3\times 2!}{2\times 1\times 2!}$$=6$

$7C_4=\large\frac{7!}{4!3!}=\frac{7\times 6\times 5\times 4!}{4!\times 3\times 2}$$=35$

$4C_1=\large\frac{4!}{1!3!}=\frac{4\times 3!}{1!\times 3!}$$=4$

$\Rightarrow 7\times 1+21\times 4+35\times 6+35\times 4$

$\Rightarrow 7+84+210+140$

$\Rightarrow 441$ ways

Hence (C) is the correct answer.

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