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How many ways are there to seat down 5 boys and 5 girls, on two parallel benches of length $5$, such that there is at least one girl opposite in front of a boy?

My attempt -

I tried to solve it by finding total ways and subtract the case when no boys and girls are opposite.

$10! - (5! \times 5!)$

But then I notice that in this problem if $4$ boys opposite to each other and then $4$ girls opposite to each other then $1$ boy and $1$ girl pending. So at least $1$ is opposite. So my above attempt is wrong. How to do this properly?

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  • $\begingroup$ Is this related to Discrete Mechatronics in any way? $\endgroup$
    – Shraddheya Shendre
    Dec 22, 2016 at 9:41
  • $\begingroup$ Also see here(but with no solution): math.stackexchange.com/questions/2068253/… $\endgroup$
    – user371838
    Dec 22, 2016 at 9:42
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    $\begingroup$ Maybe I'm confused. So you have five girls and five boys, and two benches. Any person can sit in any place, correct? I'm having difficulty imagining an arrangement where there is not at least one pair of girl and boy sitting opposite of each other. I believe all cases have at least a girl and a boy siting opposite of each other. $\endgroup$
    – Enrico Borba
    Dec 22, 2016 at 9:43
  • $\begingroup$ @Rohan - I too saw the same and I believe this is some kind of an assignment or something. $\endgroup$
    – Shraddheya Shendre
    Dec 22, 2016 at 9:43
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    $\begingroup$ There is no way in which this can happen due to odd number of boys and girls. As boys can't have a girl opposite them, there has to be a boy opposite a boy. Hence boys occur in pairs. But here number of boys is odd. So... $\endgroup$
    – Shraddheya Shendre
    Dec 22, 2016 at 9:45

2 Answers 2

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But then I notice that in this problem if $4$ boys opposite to each other and then $4$ girls opposite to each other then $1$ boy and $1$ girl pending. So at least $1$ is opposite. So my above attempt is wrong. How to do this properly?

Just do that.   You have noticed that all arrangement must have at least one girl seated opposite a boy.   That is what you needed to notice.   Hence the count you seek is the count of all possible arrangements.   Thus there is nothing to exclude; and so nothing to subtract.

The count is $~10!~$, that is all.

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As discussion cleared there is no possibility which is unfavourable

So total ways are $\frac{10!}{2}$ as benches may arrange in 2 ways

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  • $\begingroup$ It's unclear if the benches are distinct. $\endgroup$
    – Enrico Borba
    Dec 22, 2016 at 10:27

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