Seven persons are to be seated in a row.The probability that two particular persons sit next to each other is

$\begin{array}{1 1}(A)\;\large\frac{1}{3}\\(B)\;\large\frac{1}{6}\\(C)\;\large\frac{2}{7}\\(D)\;\large\frac{1}{2}\end{array}$

Number of ways in which seven persons are seated in a row =7!
Two persons sitting next to each other =6!
$\therefore$ Required probability =$\large\frac{6!}{7!}$$\times 2!$
$\Rightarrow \large\frac{2}{7}$
Hence (C) is the correct answer.