Combinatorics counting problem

Here is an alternative:

There will be $n$ groups containing $1$ male and $2$ females and the other $n$ group containing $2$ males and $1$ female.

The number of ways to divide the males into these $2n$ unordered groups is $\frac{(3n)!}{2^{n}n!n!}$.

The number of ways to divide the females into these $2n$ ordered group (yes ordered) is $\frac{(3n)!}{2^{n}}$.

Thus the number of ways to divide these males and females into group of $3$ with this requirement is $\left(\frac{(3n)!}{2^{n}n!}\right)^{2}$

I think that the answer should be $$\frac{((3n)!)^2\binom{2n}{n}}{(2n)!2^{2n}}=\left(\frac{(3n)!}{2^{n}n!}\right)^{2}.$$ Explanation: we have two lines of $3n$ persons, one with males and another for females. Then we form $2n$ groups of $3$ persons each by taking $1$ male and $1$ female from the respective lines and then we fill the rest with the tails of the lines into $\binom{2n}{n}$ ways. Finally we divide by the number of permutations of the $2n$ groups, i.e. $(2n)!$ and by $2$ for each group (since we have a pair).