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Algebra

Blazing goIITian

 Joined: 9 May 2009 Post: 766
3 Nov 2009 18:49:01 IST
0 People liked this
4
530
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.

Blazing goIITian

Joined: 5 Sep 2008
Posts: 953
3 Nov 2009 19:34:53 IST
0 people liked this

Is it 55/ 72 by any chance ?

Blazing goIITian

Joined: 14 May 2009
Posts: 697
3 Nov 2009 19:53:34 IST
5 people liked this

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

Blazing goIITian

Joined: 9 May 2009
Posts: 766
3 Nov 2009 21:20:34 IST
0 people liked this

Thanks Deepak ! The correct answer is 1/6.

Blazing goIITian

Joined: 6 May 2008
Posts: 386
19 Nov 2009 12:33:16 IST
0 people liked this

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

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