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it's the question of math&stat, the way I think is 12! (no restrictions) - the permutations that 2 girls sitting together. But there are repeated cases, and i've puzzled and I realized that there're many experts here, so I hope my problem can be solved, thanks.

the answer is 3XXXXXXX

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Could you say a little about the context of this question? How do you know the first digit? –  Noah Snyder Dec 16 '12 at 14:20
    
it's the hint.. –  Wong Kim Dec 16 '12 at 14:22
    
Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. If this is homework, please add the homework tag; people will still help, so don't worry. –  Noah Snyder Dec 16 '12 at 14:28
    
Thank you. :) i understood. –  Wong Kim Dec 16 '12 at 14:36

2 Answers 2

$\newcommand{\boy}{\color{blue}{\boxplus}}\newcommand{\girl}{\color{red}{\oplus}}$ Hint: Think of 7 generic boys first seated with a slot between each adjacent pair (and also slots on either end) as such: $$ \underline{\;} \quad \boy \quad \underline{\;} \quad \boy \quad \underline{\;} \quad \boy \quad \underline{\;} \quad \boy \quad \underline{\;} \quad \boy \quad \underline{\;} \quad \boy \quad \underline{\;} \quad \boy \quad \underline{\;} $$ Find the number of ways to put 5 generic girls in these slots (at most one per slot!). An example of doing this is $$ \girl \quad \boy \quad \underline{\;} \quad \boy \quad \girl \quad \boy \quad \girl \quad \boy \quad \underline{\;} \quad \boy \quad \girl \quad \boy \quad \girl \quad \boy \quad \underline{\;} $$ which gives the sitting arrangement $$ \girl \quad \boy \quad \boy \quad \girl \quad \boy \quad \girl \quad \boy \quad \boy \quad \girl \quad \boy \quad \girl \quad \boy $$

Now give names to each of the boys and each of the girls (by which I mean permute the boys among themselves, and permute the girls among themselves).

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Two girls that are siting together we can see as one girl they can sit in two ways.Two girls out of 5 we can choose in $\binom{5}{2}$ and all diferent positions are $$2\cdot\binom{5}{2}\frac{(7+4)!}{7!4!}=6600$$

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thx. but could u help me for the next steps? :) I've no idea. –  Wong Kim Dec 16 '12 at 14:50

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