Probability - Selection from Circular Table

Out of n persons sitting at a round table, the persons, A, B and C are selected at random. Prove that the chance that no two of these are sitting together is $\dfrac{(n – 4)(n – 5)}{(n – 1) (n – 2)}$

My thoughts: The total possible outcomes should be m(m-1)(m-2), which represents selection of any three people from the table.

In the numerator, that is, favourable outcomes, we've to find the possible ways of selecting 3 individuals who are not adjacent to each other. For the first individual, there are $n$ options, for the second, $n-3$ options, and for the third? (I'm not sure how many)

Is there a better way of getting around this? Looking at the answer, it feels like the numerator is n(n-4)(n-5), but we can't be sure.

$3$ people can be chosen from $n$ in $n(n-1)(n-2)$ ways

Let seats be numbered from $1$ to $n$.

The first person can be chosen in $n$ ways. Let us say we pick $1$. Second in $n-3$ ways.

Now if the second person could be either $3$ or $n-1$. In these cases the third can be chosen in $n-5$ ways.

if the second person is not $3$ or $n-1$, the third can be chosen in $n-6$ ways.

So total ways = $n(2(n-5) + (n-5)(n-6)) = n(n-4)(n-5)$.

So probability $= \frac{n(n-4)(n-5)}{n(n-1)(n-2)} = \frac{(n-4)(n-5)}{(n-1)(n-2)}$