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

Options

A. `3//8`
B. `1//6`
C. `3//5`
D. none of these

Answer

B

Explanation

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 speaking order is `.^(8)C_(3)`. There are 5! Ways in which the other five can speak.
So, Favourable number of ways `=.^(8)C_(3)xx5!`
Hence, required probability `=(.^(8)C_(3)xx5!)/(8!)=(1)/(6)`