$\begin{array}{1 1}(A)\;400\\(B)\;700\\(C)\;1050\\(D)\;2100\end{array} $

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

Total no. of students =10

Total no. of teams =3

Required no. of ways =$\large\frac{10C_4\times 6C_3\times 3C_3}{2!}$

$10C_4=\large\frac{10!}{4!6!}$

$\Rightarrow \large\frac{10\times 9\times 8\times 7\times 6!}{4\times 3\times 2\times 6!}$$=210$

$6C_3=\large\frac{6!}{3!3!}$

$\Rightarrow \large\frac{6\times 5\times 4\times 3!}{3!0!}$$=1$

$\Rightarrow \large\frac{210\times 20\times 1}{2}$

$\Rightarrow 2100$

Hence (D) is the correct answer.

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