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

Dear Student,

For 8 speakers, total arrangement will be 8!. Now out of 8 postions selecting any three positions for (A,B and C in order) will be ⁸C₃ are remaining can be arranged in 5! ways.

Therefore, Probability will be (⁸C₃)*5!/8! = 1/6

Thanks.