Recursive and closed form solution for choosing $n$ pairs/triplets.. of $kn$ elements. I stumbled apon an interesting question:
How many ways are there to arrenge $kn$ elements into $n$ sets, $k$ elements each?
There should be a recursive and closed  form solution for $g_k(n)$.
For example, the closed form for $k=2$ (arrenge $2n$ elements into pairs) is: $$1\cdot3\cdot5 ... \cdot(2n-1)$$
And the recursive solution for $k=3$ (arrenge $3n$ elements into triplets) is: $$g_3(n)={{3n-1}\choose 2}g_3(n-1)$$
(I can't figure out how to get to those either)
 A: For closed form, do the division as follows: line up the $kn$ people in a row, put the first $k$ into one group, the next $k$ into another, and so on.
There are $(kn)!$ ways to line up the $kn$ people, but this grossly overcounts the number of divisions into groups.
Any of the $n!$ permutations of the chunks gives the same division into groups. And within any chunk of size $k$ we can permute the $k$ people arbitrarily without latering the division into groups. Thus the "overcount" factor is $n!(k!)^n$, and the number of ways to divide the $kn$ people into $n$ groups of $k$ is $\dfrac{(kn)!}{n!(k!)^n}$.
Remark: There is a simple way to visualize the result you mention for $k=2$. Lin up the people in order of height, or beauty, or student number.
The first person has $2n-1$ ways to choose here partner. That leaves $2n-2$ people. The first unchosen person has $2n-3$ ways to choose her partner. That leaves $2n-4$ people. The first unchosen person has $2n-5$ ways to choose her partner. And so on. Thus the total number of choices is $(2n-1)(2n-3)(2n-5)\cdots(3)(1)$.
You might want to see how the formula given in the main answer gives the same number in the case $k=2$. 
