Question

Three persons $A1,A2$ and $A3$ are to speak at a function along with $5$ other persons. If the persons speak in random order, the probability that $A1$ speaks before $A2$ and $A2$ speaks before $A3$ is

• A. $\frac{1}{6}$
• B. $\frac{3}{5}$
• C. $\frac{3}{8}$
• D. None of these

Answer 1

The correct option is A $\frac{1}{6}$

For total 8 speakers, total arrangement will be 8!. Now out of 8 positions selecting any three positions for (A,B and C in order) will be ${}^8{C}_3$ are remaining can be arranged in 5! ways.

Therefore, Probability will be ${(}^8{C}_3)\times \frac{5!}{8!}=\frac{1}{6}$