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Q)

In how many number of ways can 10 students be divided into three teams,one containing four students and the other three?

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

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A)
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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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