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:

1/6

Total ways = 8!

Favorable outcomes = (choose '3' places for ABC) × arrage rest 5 people = ${^8C}_3 . 5 !$

P(req.) = $\frac{({^8C}_3 × 5!)}{8!} =\frac{1}{6}$