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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?

  • Is this related to Discrete Mechatronics in any way? – Shraddheya Shendre Dec 22 '16 at 09:41
  • Also see here(but with no solution): http://math.stackexchange.com/questions/2068253/discrete-mechatronics-sequences-with-repeats-and-no-repeats –  Dec 22 '16 at 09:42
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    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. – Enrico Borba Dec 22 '16 at 09:43
  • @Rohan - I too saw the same and I believe this is some kind of an assignment or something. – Shraddheya Shendre Dec 22 '16 at 09:43
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    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... – Shraddheya Shendre Dec 22 '16 at 09:45
  • @Rohan if something is tricky. I think nothing wrong in asking that question with more details. –  Dec 22 '16 at 09:47
  • @ShraddheyaShendre Exactly. So now it's just a matter of counting the number of arrangements of 5 elements of one kind and 5 elements of another. – Enrico Borba Dec 22 '16 at 09:47
  • Are the girls distinct? Are the boys distinct? – Enrico Borba Dec 22 '16 at 09:48
  • Yes 5 girls are distinct. –  Dec 22 '16 at 09:54

2 Answers2

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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.

Graham Kemp
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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

Blaise Thunderstorm
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