There are 8 stations between two cities A and B. A train is to stop at three of these 8 stations. The probability that no two of these three stations are consecutive is

Let A be the required event and S be the sample space. There are 8 stations of which 3 are halting and the remaining 5 are non halting stations.
Now $\mathrm{n}\left(\mathrm{S}\right)=8!$
Since no two halting stations are consecutive, first we arrange the 5 non halting ${}^6\mathrm{P}_3$ stations. It can be done in 5! Ways.
Now there are 6 places to arranged the 3 halting stations. They can be arranged in   ways.
Thus, $\mathrm{n}\left(\mathrm{A}\right)=5{!}^6{\mathrm{P}}_3$
$\therefore (\mathrm{P}\left(\mathrm{A}\right)=\frac{\mathrm{n}\left(\mathrm{A}\right)}{\mathrm{n}\left(\mathrm{S}\right)}=\frac{5{!}^6{\mathrm{P}}_3}{8!}=\frac{6\times 5\times 4}{8\times 7\times 6}=\frac{5}{14}$