Three persons A, B and C are to speak at a function along with five others. If they all speak in random order, the probability that A speaks before B and B speaks before C is

$\frac{1}{6}$

The total number of ways in which 8 persons can speak is ${^8P_8} = 8!$ The number of ways in which A, B and C can be arranged in the specified speaking order is ${^8C_3}$. There are 5! ways in which the other five can speak.

So, Favourable number of ways $={^8C_3}× 5!$

Hence, required probability $= \frac{^8C_3× 5!}{8!}=\frac{1}{6}$