20 recruits distributed into 4 groups each consisting of 5 recruits 20 recruits distributed into 4 groups each consisting of 5 recruits. In how many ways can we do that?
The answer is $\frac{20!}{(5!)^4\cdot4!}$, However I do not understand why can't we just do the following:
We have $4$ groups, in each we can have $5!$ of distributions of recruits, and out of those $4$ groups we can also do $4!$, so why isn't the answer $4!\cdot 5!$ ?
 A: First of all, if there are $5!$ possibilities to line up the recruits for each of the $4$ groups, then you should get $4! \cdot (5!)^4$ possibiltiies.
OK, but that's still not right:
$4! \cdot (5!)^4$ is the number of possibilities of lining up all 20 recruits, where order matters, and where apparently you have already decided the groups.  That is, if you already have created the $4$ groups of $5$, and you now line them up, but in groups of $5$ (i.e. the first $5$ in the line-up belong to one group, numbers 6 through 10 belong to a second group, etc., then indeed you get $4!$ (the number of ways you can order to the groups in the line-up) times $(5!)^4$ (the number of ways to line up the $5$ members within a group, for all of the $4$ groups)
However, the question wasn't to line them up: it was simply to divide the recruits into groups. So, a group consisting of members $3,6,7,12$, and $17$ is the same group as the group consisting of $6,12,17,3$, and $7$. Likewise, you are not ordering the groups at all: there is no 'first' group of 'second' group .. there are just $4$ groups total.
Also, as I said, you assume that you already have $4$ groups of $5$ recruits, but the question was to figure out the number of ways to find those $4$ groups of $5$ recruits.
So, if you line up all $20$ recruits, and then say: the first $5$ in line forms group one, the next $5$ are group two, etc. then you going into the right direction.  Now, there are $20!$ ways to do that, but you are overcounting, again because order does not matter. That is: if in the line-up the first $4$ had switched position with the second $5$, you would have ended up with the same groupings. That is why we divide by $4!$. And likewise: had the $5$ members of a group switched posisions, it would have remoined that same group. So that is why we also divide by $5!$ for each group, i.e. we divide by $(5!)^4$. So, there are:
$$\frac{20!}{4!(5!)^4}$$
ways to find those $4$ groups of $5$
