For class discussion, arrange 12 students of class 12a consisting of 6 male students and 6 female students into a line consisting of 2 lines of chairs opposite to each other (6 chair each). Find the probability to have 2 students sitting opposite to and next to each other always having opposite sex.
I tried
There are $ 12! $ ways to arrange $ 12 $ students. Consider the cace as following table
The first Boy has 6 ways to seat;
The second Boy has 5 ways to seat;
The third Boy has 4 ways to seat;
The fourth Boy has 3 ways to seat;
The fifth student Boy has 2 ways to seat;
The sixth student Boy has 1 ways to seat;
6 girls has $ 6! $ ways to seat.
Therefore, with the above table, we have
$$ 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 \cdot 6!.$$
With the table
we also have
$$ 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 \cdot 6!.$$ ways.
The probility we need to find is
$$ \dfrac{2 \cdot 6!}{12!} = \dfrac{1}{462} $$
Is my solution correct?
Your revised solution is correct.
The boys can be arranged in a row in $6!$ ways. The girls can also be arranged in a row in $6!$. There are two ways to decide whether a boy or girls sits in the top left corner, which completely determines where the other boys and girls sit. Hence, there are $2 \cdot 6! \cdot 6!$ favorable arrangements. Thus, the desired probability is $$\frac{2 \cdot 6! \cdot 6!}{12!}$$