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.

By using combination, you cannot use $C(7,6)$ because this means the girls can choose any 6 from among those 7 blank seats. What if 2 of the chosen boxes were the two ends? So one of the middle boxes must not be occupied by a girl, and hence, two boys become adjacent (which you do not want to happen).
