In a conference 10 speakers are to give their speeches one after another. Find the probability of the event if S₁ speaks before S₂ and S₂ speaks before S₃ and the remaining 7 speakers have no objection to speak at any number.

 ten speaker can address in 10! ways.......including S1 S2 S3

now s1 s2 s3 can be arranged in 3! ways=6ways i.e  s1s2s3 , s1s3s2,s2s1s3,s2s3s1,s3s1s2,s3s2s1

since we want only the order s1 s2 s3 we will take 1/6 th of the total....

so no of ways such that S₁speaks before S₂ and S₂ speaks before S₃ is (1/6)10!

no of ways in which  ten speaker can address is 10! ways

probab=(1/6)10!/10!=1/6

Thanks Deepak ! The correct answer is 1/6.

 i din get that logic..........why to take 1/6 th