Doubt in high school permutation question(seating arrangement) 
$6$ boys and $6$ girls form a line with boys and girls alternating.Find the number of ways of making the line.

Answer $2(6!)^2$ 
I was trying to solve this question using Permutation and Combination 
$$\square B \square B \square B \square B \square B \square B \square$$
Blank boxes shown above are for girls.Therefore girls can be chosen in $C(7,6)$. Now boys and girls can be arranged in $6!$ ways.
Therefore, it's answer must be $C(7,6) \cdot 6! \cdot 6!$.
Please point out my mistake.
Note- I got correct answer using Permutation and logical thinking but I want to solve by Combination too.


Therefore girls can be chosen in C(7,6)

Note that C(7,6) (out of seven, pick six to include) is the same as C(7,1) (out of seven, pick one to exclude), which is just 7. There are seven boxes, so it doesn't make sense to say that this is how many ways there are of choosing girls. It would make more sense to say that this is how many ways of choosing boxes for girls. But that still isn't correct, because every box between two boys must be chosen (otherwise, you would end up with two boys next to each other). So there are only two choices for which box to exclude: either the first box can be excluded, or the last.
