Question

Three persons $A,B,C$ are to speak at a function along with $5$ other persons. If the persons speak in random order, the probability that $A$ speaks before $B$ and $B$ speaks before $C$ is $\frac{p}{q},$ where $p,q$ are co-prime. Then $p+q$ is

Answer 1

Total number of ways in which $8$ persons can speak is $8!$

Number of ways in which $A,B$ and $C$ can be arranged in the specified speak order is ${}^8{C}_3\times 1$ as the order of $A,B,C$ is already fixed.
Remaining $5$ persons can speak in $5!$ ways.
So the favourable number of ways $={}^8{C}_3\times 5!$
Hence, required probability $=\frac{{}^8{C}_3\times 5!}{8!}=\frac{1}{6}$