Number of ways to choose disjoint subsets. 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.
 A: 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 $t \rightarrow \infty$.
A: Relate this problem to the question of "In how many ways can you arrange the letters in the word $\underbrace{aa\dots a}\underbrace{bb\dots b}\underbrace{cc\dots c}_{r~\text{of each}}\dots\underbrace{tt\dots t}$"

 There will be $\binom{rt}{r,r,\dots,r} = \frac{(rt)!}{(r!)^t}$ different ways to do so.

A: Without loss of generality let $A=\{1,2,\ldots,rt\}$. Imagine filling an $r\times t$ array with the elements of $A$. There are $(rt)!$ possible ways to do so, and each gives you a partition of $A$ into $t$ sets of cardinality $r$, one for each row. Of course each of those sets could have been entered into the array in $r!$ different orders, so there are $(r!)^t$ different filled arrays that correspond to the same partition of $A$. Thus, the number of such partitions is
$$\frac{(rt)!}{(r!)^t}=\binom{rt}{\underbrace{r,r,\ldots,r}_t}\;.$$
A: The formula part was easy:
$$\frac{(rt)!}{r!^t}$$
I went looking for the name ... "Multi-select combination"?
