# Sitting arrangement problem.

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? Dec 22, 2016 at 9:41
• Also see here(but with no solution): math.stackexchange.com/questions/2068253/…
– user371838
Dec 22, 2016 at 9:42
• 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. Dec 22, 2016 at 9:43
• @Rohan - I too saw the same and I believe this is some kind of an assignment or something. Dec 22, 2016 at 9:43
• 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... Dec 22, 2016 at 9:45

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?
The count is $~10!~$, that is all.
So total ways are $\frac{10!}{2}$ as benches may arrange in 2 ways