In how many different ways can boys and girls sit a desks such that at each desk only one girl and one boy sits? There are $n$ boys, $n$ girls and $n$ desks. In how many different ways can boys and girls sit a desks such that at each desk only one girl and one boy sits?
 I have a solution, but I have a little doubt that it is incomplete. So my solution is as follows:

My worked solution: For the first desk  we can choose one girl by $\binom n1$ and one boy by $\binom n1$ and we can permute them by $2!$ ways. In result we have $2!\binom n1 \binom n1$. In the same way, we can work out for the second desk $2!\binom {n-1}1 \binom {n-1}1$ and so on. For the $nth$ desk we have $2!\binom {1}1 \binom {1}1$ ways.

 $Total = 2!\binom n1 \binom n1 2!\binom {n-1}1 \binom {n-1}1 ... 2!\binom {1}1 \binom {1}1 = 2^n(n!)^2$

To remove from unanswered queue:
Yes, you are correct.  Approaching via multiplication principle, first choose whether the boy is on the left, or the right for each desk in sequence.  Then in sequence, choose which boy is at the desk and which girl is at the desk for a final total of:
$$2^n(n!)(n!)$$
