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A team of four students is to be selected from a total of 12 students.Total number of ways in which team can be selected such that two particular students refuse to be together and other two particular students wish to be together only is equal to

$\begin{array}{1 1}(A)\;220\\(B)\;182\\(C)\;226\\(D)\;\text{None of these}\end{array} $

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  • $n(C,r)=\large\frac{n!}{r!(n-r)!}$
Let $S_1$ and $S_2$ refuse to be together and $S_3$ and $S_4$ want to be together only.
Total ways when $S_3$ and $S_4$ are selected
$\Rightarrow 8C_2+2C_1+8C_1$
$8C_2=\large\frac{8!}{6!\times 2!}=\frac{8\times 7\times 6!}{2\times 6!}$
$\Rightarrow 28$
$\Rightarrow 44$
Total ways when $S_3$ and $S_4$ are not selected =$(8C_2+2C_1.8C_3)$
$\Rightarrow 182$
Thus total ways =$44+182$
$\Rightarrow 226$
Hence (C) is the correct answer.
answered May 19, 2014 by sreemathi.v

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