There are n stations between two cities A and B. A train is to stop at three of these n stations. What is the probability that no two of these three stations are consecutive ?

$\frac{(n-3)(n-4)}{n(n-1)}$

The total number of ways of choosing 3 stations out of n stations is ${^nC}_3.$

So, total number of ways = ${^nC}_3.$

Let $x_1$ be the number of stations before the first halting station, $x_2$ between first and second, $x_3$ between second and third and $x_4$ on the right of third station. Then, $x_1 ≥ 0, x_2 ≥ 1, x_3 ≥1$ and $x_4 ≥ 0$ such that $x_1 + x_2 + x_3 + x_4 =n-3$.

The total number of solutions of this equation is ${^{n-2}C}_3$.

Hence, the required probability $=\frac{^{n-2}C_3}{^nC_3}=\frac{(n-3)(n-4)}{n(n-1)}$