Sitting arrangement problem.

Question

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?

Answer 1

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.

Answer 2

As discussion cleared there is no possibility which is unfavourable

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