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
Thanks Deepak ! The correct answer is 1/6.
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