$\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

$\Rightarrow 8C_2+2C_1+8C_1$

$8C_2=\large\frac{8!}{6!\times 2!}=\frac{8\times 7\times 6!}{2\times 6!}$

$\Rightarrow 28$

$2C_1=\large\frac{2!}{1!1!}$$=2$

$8C_1=\large\frac{8!}{1!7!}$$=8$

$8C_2+2C_1.8C_1=28+16$

$\Rightarrow 44$

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.

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