Hi NandishSS,
This question can be approached in a number of different ways (depending on how you choose to do the math involved). It's far easier to calculate the probability that NO girls will sit next to one another, so I'm going to calculate that and then subtract that fraction from the number 1; this will give us the probability that two (or more) girls will end up sitting next to one another.
To start, any of the 10 chairs can be the 'first chair' - and any of the 10 people can sit in it. Once someone is placed in that seat, we want the next seat have someone of the opposite sex (and there are 5 options for that 'second chair'). From here, the pattern is now set for the other 8 chairs (it's either BGBGBGBG or GBGBGBGB, depending on who sits in the 'first chair') - and we'll end up with 4 possible kids for the 'third chair', 4 for the 'fourth chair', 3 for the 'fifth chair', 3 for the 'sixth chair', etc.
This means that there are (10)(5)(4)(4)(3)(3)(2)(2)(1)(1) possible ways to arrange the boys and girls in every-other-seat. Do NOT multiply that out yet.
The total possible ways to place the 10 people in the 10 seats is 10!, which I'm going to expand-out in a moment. We now have our probability:
(10)(5)(4)(4)(3)(3)(2)(2)(1)(1) / (10)(9)(8)(7)(6)(5)(4)(3)(2)(1)
You can 'cancel out' a LOT of the numbers in the numerator and denominator. For example, the 10s 'cancel out', you can take the two 3s from the numerator and 'cancel out' the 9 in the denominator, etc. By doing that, I was left with:
(1) / (7)(6)(3) = 1/126
Thus, since there's a 1/126 probability of having NO girls sit next to one another, there's a 1 - (1/126) = 125/126 probability of having 2 (or more) girls sit next to each other.
Final Answer:
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