Question

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

• A. $3/8$
• B. $1/6$
• C. $3/5$
• D. None of these

Answer 1

The correct option is B $1/6$

The total number of ways in which $8$ persons can speak is ${}^8{P}_8=8!$
The number of ways in which $A,B$ and $C$ can be arranged in the specified speak order is ${}^8{C}_3.$
There are $5!$ ways in which the other five can speak .
So the favorable number of ways is ${}^8{C}_3=S!$
Hence, the required probability $\frac{{}^8{C}_3\times S!}{8!}=\frac{1}{6}.$