# Find the probability to have 2 students sitting opposite to and next to each other always having opposite sex.

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?

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 $$\frac{2 \cdot 6! \cdot 6!}{12!}$$