Combinatorics counting problem

Question

There are $3n$ male students and $3n$ female students. How many ways can they be divided into two groups of three, in such way that in each group has at least one male student and one female student.

I'd like to know if the following is correct: First I organize them in lines of $2n$, so: $2n!\cdot 2n!$. then, I create a third line with the rest, so in total: $(2n!)^3$. But now, I have created order inside the triplets, so to remove it I divide by $(3!)^{2n}$.

The final solution should be: $$\frac{(2n!)^3}{(3!)^{2n}}.$$

Is the solution above correct? Is the way of thinking correct?

Answer 1

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).

Answer 2

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 $\left(\frac{(3n)!}{2^{n}n!}\right)^{2}$