Plz solve diss..

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Plz solve diss..

Jaspreet kaur
If n people are seated in a random manner in a row
containing 2n seats, what is the probability that no two
people will occupy adjacent seats?
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Re: Plz solve diss..

cheesecake
This may be completely wrong.
Total ways in which n people can be seated on 2n chairs will by 2nPn. Now let the first persob be seated on chair 1. 2nd person on chair 3, 3rd person on chair 5 and so on. This can be done in n! ways. Also it is possible that the 1st person was seated on chair 2 instead of chair 1, 2nd person on chair 4 and so on. This happens in n! Ways too. So total number of ways is n!+n! = 2(n!).
So probability is 2(n!) / 2nPn
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Re: Plz solve diss..

Jaspreet kaur
The answer is n+1 ways/ nC2
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Re: Plz solve diss..

ABHI1994
for cross check solution take n=2 and solve,u will get ans.

On 24 June 2015 at 04:15, Jaspreet kaur [via Discussion forum] <[hidden email]> wrote:
The answer is n+1 ways/ nC2


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Re: Plz solve diss..

Jaspreet kaur
In reply to this post by Jaspreet kaur
Let n=2 nd 2n=4 ,following r the arrangements

0_0_ or _0_0
 And also
0_ _ 0 ...so numerator will be what?
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Re: Plz solve diss..

amianand
My interpretation of question is that none of the persons are sitting on adjacent seats. Though it can also be that no 2(particular 2) are not sitting together. In that way question is not clear. Anyways, Consider, you have to select n seats from 2n seats such that none of selected seats are together. So, you dont have to select n seats, there will be n+1 gaps between these n seats. Select n gaps out of these n+1 gaps and place your remaining n seats here. i.e. C(n+1,n).n!/C(2n,n).n! = (n+1)/C(2n,n)
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Re: Plz solve diss..

amianand
In reply to this post by Jaspreet kaur
My interpretation of question is that none of the persons are sitting on adjacent seats.
Though it can also be that no 2(particular 2) are not sitting together. In that way question is not clear.

Anyway,

Consider, you have to select n seats from 2n seats such that none of selected seats are together.

So, you dont have to select n seats, there will be n+1 gaps between these n seats.

Select n gaps out of these n+1 gaps and place your remaining n seats here.

 i.e.

C(n+1,n).n!/C(2n,n).n!

= (n+1)/C(2n,n)
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Re: Plz solve diss..

Jaspreet kaur
Gt it :)
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Re: Plz solve diss..

cheesecake
How are there n+1 gaps?
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Re: Plz solve diss..

cheesecake
In reply to this post by Jaspreet kaur
Nevermind i got it.. Thanks amianand! The way i did it, i was completely ignoring the cases where there might be more than one space between two people! :P