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suppose we have two rows and each contain 5 seats and we have 5 girls(girl 1,2,3,4,5) and 5 boys(boy 1,2,3,4,5) What are the number of cases that all boys sat together or girl 1 and 2 sat next to each other?

This is what I have done: $5!\cdot5!\cdot2+8\cdot8!\cdot2=\text{answer}$ But I am 100% sure I did over counting, my question in how to fix those over counting.

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    For example there is a case such that in first row there are all women and in second raw all men, 2 in both cases in for first or second row and 5! Is for positing of girl and other 5! Is for positing boys and 8 is for posing two girls next to each other and 8! For other reaming boys or girls2012-12-03

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Yes, you have overcounted. To correct for it, you must subtract the number of arrangements that you’ve counted twice. These are the arrangements that have all of the boys in one row, all of the girls in the other row, and girls $1$ and $2$ next to each other. There are $2\cdot5!\cdot4\cdot2\cdot3!$ of these arrangements: a choice of $2$ rows for the boys, $5$ ways to permute the boys, a choice of $4$ positions for the pair of girls, a choice of $2$ orders for girls $1$ and $2$ within that pair, and a choice of $3!$ permutations of the other $3$ girls. Your final answer should therefore be

$5!\cdot5!\cdot2+8\cdot8!\cdot2-2\cdot5!\cdot4\cdot2\cdot3!\;.$

This an example of the method of inclusion-exclusion.

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    @Hooman: Thank you; you’re very welcome.2012-12-03