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
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)
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.
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