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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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$\newcommand{\boy}{\color{blue}{\boxplus}}\newcommand{\girl}{\color{red}{\oplus}}$ 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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