Combination/Permutation Dilemma This is probably quite a straight forward one - but still struggling to get my head around it. 
To explain my problem:
In my work, I'm trying to put subjects into groups of 4 for a second round of an experiment based on their test scores in the first round (n = 20), therefore 5 groups in total. 
There are 20 C 4 ways of combining the individuals into groups of size 4 = 4845 possible different groups. What I want to know is how many unique combinations are possible of 5 different groups of 4, given that individuals are free to vary within each.
The applied purpose of this is that I am writing a code that allocates individuals into groups so as to minimise the between groups mean and variances, so need to know how many iterations to let the simulation run through.
To embark on a total digression, upon writing this post title I also discovered 'dilemma' is not in fact spelled 'dilemna' (as is apparently also widely believed by many) so am already 1 up today. 
 A: Denote the groups 1,2,3,4,5. First we pick 4 individuals for group 1. That gives $\binom{20}{4}$ ways. Now we pick 4 individuals for group 2: $\binom{16}{4}$. For group 3: $\binom{12}{4}$. This gives:
$$\binom{20}{4}\binom{16}{4}\binom{12}{4}\binom{8}{4}\binom{4}{4}=\frac{20!}{4!^5}.$$
However this doesn't take into account multiplicity, so it depends on what you're asking. In the above formulation each group is different. So if Alice, Bob, Sam and Jenny go into group 1, that would be a different combination then if Alice, Bob, Sam and Jenny go into group 4. If you'd like to not distinguish between group numbers, you'll need to divide by an extra $4!$ to account for the group orderings. In the latter interpretation, the only thing that matters is that Alice, Bob, Sam and Jenny are in the same group, regardless of it's number.
A: One approach: 20! ways to order the people. But the order within each group doesn't matter; so divide by $4!^5$. Finally, the order of the groups doesn't matter, so divide by 5!.
The answer is then $\frac{20!}{5!4!^5} = 2546168625$.
To divide $n*k$ people into $n$ groups of $k$ is then $\frac{(n*k)!}{n!k!^n}$.
(Compared to the approach/assumptions here; if one finishes by saying that the five groups are indistinguishable one divides by 5! and the same result is achieved.)
