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Blazing goIITian

Joined:	9 May 2009
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3 Nov 2009 18:49:01 IST

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Probability

Engineering Entrance , JEE Main , JEE Advanced , Mathematics , Algebra

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.

Comments (4)

AViK

Blazing goIITian

Joined: 5 Sep 2008

Posts: 953

3 Nov 2009 19:34:53 IST

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Is it 55/ 72 by any chance ?

eragon24

Blazing goIITian

Joined: 14 May 2009

Posts: 697

3 Nov 2009 19:53:34 IST

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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

Deepak Aggarwal

Blazing goIITian

Joined: 9 May 2009

Posts: 766

3 Nov 2009 21:20:34 IST

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Thanks Deepak ! The correct answer is 1/6.

akl

Blazing goIITian

Joined: 6 May 2008

Posts: 386

19 Nov 2009 12:33:16 IST

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

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