Given a set $A$ of cardinality $rt$ where $r,t\in\Bbb N$ how many ways can you choose $t$ disjoint subsets of cardinality $r$? Is there an elementary formula and is there a name for this problem?
I am also looking for asymptotic solutions for case $t$ is fixed and for case both $r$ and $t$ grow according to some function.
Best Answer
As the other answers have stated, if you want to partition $A$ into pairwise-disjoint sets $A_1, A_2, ..., A_t$, each of size $r$, there are $\binom{rt}{\underbrace{r,r,...,r}_{t \text{ times}}} = \frac{(rt)!}{(r!)^t}$.
One way to see where this formula comes from is to imagine ordering all $rt$ elements ($(rt)!$ possibilities), and then taking the first $r$ to make the first set, the second $r$ to make the second set, and so on, until the last ($t$th) set of $r$ elements form the final set. However, the relative order of the elements in each set is irrelevant, so we should divide through by a factor of $r!$ for each of the $t$ sets.
Note that in this answer, the sets themselves are ordered - we have a first set $A_1$, a second set $A_2$, and so on, until a $t$th set $A_t$. If you do not care about the order of the sets, then you should further divide by the $t!$ ways there are of ordering the sets, which would give an answer of $\frac{(rt)!}{(r!)^t t!}$.
Finally, the asymptotic analysis will depend on how our parameters behave. Note that Stirling's Approximation gives $n! = (1 + o(1)) \sqrt{ 2 \pi n} \left( \frac{n}{e} \right)^n$ when $n \rightarrow \infty$. If, however, $n$ is fixed, then one must use $n!$ in asymptotic estimates ($n!$ is just a constant).
Hence, if $r$ is fixed and $t \rightarrow \infty$, $$ \binom{rt}{r, r, ..., r} = \frac{(rt)!}{(r!)^t} = \frac{ (1 + o(1)) \sqrt{2 \pi rt} \left( \frac{rt}{e} \right)^{rt} }{ (r!)^t} = (1 + o(1)) \sqrt{ 2 \pi rt} \left( \frac{r^r t^r}{r! e^r} \right)^t. $$
If $r$ tends to infinity as well, then we can also estimate the $r!$ in the above expression, giving $$ \binom{rt}{r, r, ..., r} = (1 + o(1)) \sqrt{2 \pi rt} \left( \frac{t^r}{( 1 + o(1)) \sqrt{ 2 \pi r} } \right)^t = \left( \frac{(1 +o(1)) t^r}{ \sqrt{ 2 \pi r} } \right)^t, $$ since the $\sqrt{2 \pi rt}$ is insignificant compared to $t^{rt}$.
Note that the latter estimate is valid even when $r \rightarrow \infty$ but $t$ is fixed, since then we still have $rt$ and $r$ tending to infinity, and hence our use of Stirling's Approximation is valid. In this case, though, since $(1 + o(1))^t = (1 + o(1))$, we can take the error term from inside the parentheses outside and get the sharper expression $(1 + o(1)) \sqrt{2 \pi r t } \left( \frac{t^r}{\sqrt{2 \pi r}} \right)^t$.
Finally, if the order of the sets is unimportant, then you should divide all of the above estimates by $t!$, which you can approximate by $\sqrt{2 \pi t} \left( \frac{t}{e} \right)^t$ when $t \rightarrow \infty$.