probability that $2$ perticular persons $A$ and $B$ are not next to each other. $n$ persons sit a round table randomly. Find the probability that $2$ perticular
persons $A$ and $B$ are not next to each other.

Attempt: Total ways of arranging $n$ persons sit around a table is $(n-1)!$
probability that $2$ perticular persons are always together is $\displaystyle \binom{n}{2}\cdot (n-1)$
So desired probability is $\displaystyle \frac{1}{(n-1)!}\times \binom{n}{2}\cdot (n-1)$
But anser mention is $\displaystyle \frac{n-3}{n-1}$, could some help me
Due to rotational symmetry, place $A$ at "12 o'clock." 

There are $n-3$ places $B$ can be placed, out of a total of $n-1$ positions.
First, let other $(n-2)$ persons sit around the table in $(n-3)!$ ways, then A and B can sit anywhere in the $(n-2)$ gaps between $(n-2)$ persons, but they cannot sit in the same gap. So, first choose two gaps in $^{n-2}C_2$ ways and then you may choose in which of these two gaps A has to sit in $2$ ways. So the final answer is $(n-3)!* ^{n-2}C_2 *2=\dfrac{(n-3)!(n-2)!}{2!(n-4)!}*2=(n-3)*(n-2)!$

Alternative answer:
Total number of ways to make $n$ people sit around the table is $(n-1)!$ ways. But we have to subtract those cases where A and B sit together, but then as they are always together we can think of them as a single person so that it is same as $(n-1)$ people sitting around the table that can happen in $(n-2)!$ ways. But the order of A and B while sitting together can be chosen in 2 ways. So we should actually subtract $2*(n-2)!$. Therefore the final answer is $(n-1)!-2*(n-2)!=(n-3)*(n-2)!$