Answer: $\mathbf B$
Explanation:
The number of integers are $\mathbf n$
Total number of arrangements possible $ = \mathbf n!$
Now, let's bundle $1$ and $2$ together, then number of arrangements possible $\mathrm{=(n-1)!2!}$ ways.
$\because$ Total elements now are $\color {blue}{\mathrm n-1}$ and one bundle is having $2$ elements which can be arranged in $2!$ ways.
So, favorable cases $\mathrm{=(n-1)!2!}$
So, total cases possible, i.e., sample space $=\mathrm n!$
$\therefore$ Probability $\large{\mathrm{= \frac{(n-1)!2!}{n!} = \frac{(n-1)!2!}{(n-1)!n}=\color {green} {\frac{2}{n} }}}$
$\therefore \mathbf B$ is the correct option.