Partitions of a multiset into blocks of same size I've got a combinatorial question: 
Let $n, k\ge 1$ integers and let $S$ be a finite multiset of order $nk$ (counted with multiplicities). I want to count partitions of $S$ into blocks subject to the following conditions: 
(1) Each block has got exactly k elements (counted with multiplicities)
(2) The order of the blocks in the partition is irrelevant, i.e. permuting the blocks inside the partition doesn't give rise to a new partition  
(3) The order of the elements in a block is relevant, i.e. permuting the elements in a block can give rise to a new block  
Example: $S=\{1,2,2,3\}$, $n=k=2$. Then $\{\{1,2\},\{2,3\}\}$ is such a partition and $\{\{2,3\},\{1,2\}\}$ is the same one but $\{\{1,2\},\{3,2\}\}$ is a different one.
 Counting this way we get six partitions:
 $\{\{1,2\},\{2,3\}\}, 
\{\{1,2\},\{3,2\}\}, \{\{1,3\},\{2,2\}\}, \{\{2,1\},\{2,3\}\}, \{\{2,1\},
\{3,2\}\}, \{\{2,2\},\{3,1\}\}$ 
Is there a name for this kind of partition? 
Does anybody know a formula for the number of such partitions? Asymptotics? 
Clearly, if all elements in $S$ are different the searched for formula is $(nk)!/n!$. But what if there are multiple elements?
 A: The approach noted in the Question of computing $(nk)!/n!$, which is exact if all the items are distinct, in general gives an upper bound on the counting.  A modest change in computation gives us a lower bound:

Let the multiplicities of distinct items be $m_1,m_2,\ldots,m_t$, so that: $$ \sum_{i=1}^t m_i = nk $$ Then the desired count will be at least: $$\frac{(nk)!}{m_1!m_2!\ldots m_t! n!} $$

One way to realize this is a lower bound is by choosing an arrangement of all $nk$ items.  By the multinomial formula there are $ \binom{nk}{m_1,m_2,\ldots,m_t}=\frac{(nk)!}{m_1!m_2!\ldots m_t!} $ ways, which might be thought of as the "single block" case $n=1$.  Now snip the single block into a sequence of $n$ equal sub-blocks.  If all the blocks were distinct, then dividing by $n!$ gives the exact corresponding number of arrangements, but since in general there could be identical sub-blocks, this can be an over-reduction.  Thus the above is only a lower bound.
In the example provided in the Question this does give us an exact count:  
$$ \frac{4!}{1!2!1!2!} = 6 $$
because it is impossible with those multiplicities to get two identical sub-blocks.
This approach can be modified (with additional effort required) to give an exact count.  We introduce some terminology to aid in formulation.
Definition Let $\mathscr{S}$ be a multi-set of $k$-permutations drawn from multi-set $A$ whose separate multiplicities add up to the corresponding total multiplicities in $A$.  We will say that $\mathscr{S}$ is a $k$-tuple resolution of $A$, but this is just to coin a phrase.
For each $k$-resolution $\mathscr{S}$ of $A$, we associate with it $\sigma_\mathscr{S}$ an integer partition of $n$ given by counting the multiplicities of the $k$-permutations belonging to $\mathscr{S}$.  If all the $k$-permutations in $\mathscr{S}$ are distinct, then $\sigma_\mathscr{S}$ is simply the sum-of-ones $1+1+\ldots+1=n$.  Otherwise the partition $\sigma_\mathscr{S}$ will have some part greater than $1$.
Let $\Sigma = \sigma_\mathscr{S}$ be a random integer partition of $n$ defined by randomly sampling $\mathscr{S}$ as outlined in the above, splitting a random permutation of $A$ into a sequence of $k$-permutations, which are then collected into the multi-set $\mathscr{S}$.
Let $\mu(\sigma) = \frac{n!}{p_1!p_2!\ldots p_s!} = \binom{n}{p_1,p_2,\ldots,p_s}$ be the multinomial coefficient corresponding to integer partition $\sigma$ of $n$ with $s$ parts:
$$ p_1 + p_2 + \ldots + p_s = n $$
If $\mathbf{P_n}$ is the set of all integer partitions of $n$, the probability distribution of $\Sigma$ determined by the above sampling produces probability values $p_\sigma = \Pr(\Sigma=\sigma)$ for each $\sigma \in \mathbf{P_n}$.
The exact count of distinct $k$-resolutions of $A$ is then:
$$ \frac{(nk)!}{m_1!m_2!\ldots m_t!}  \sum_{\sigma \in \mathbf{P_n} } \frac{p_\sigma}{\mu(\sigma)} $$
The difficulty in computing the probabilities $p_\sigma$ depends on the multiplicities $m_i$ of items in $A$.  When $k$ is large and a high fraction of the items belong to the classes of items with small multiplicities, then the probability $p_\sigma$ will be concentrated on the sum-of-ones integer partition $\sigma$.  Then our lower bound above will be correspondingly close to the actual value.
References
I'm unable as yet to give a citation that considers the OP's exact problem.  It concerns "partitioning" a multiset, not into sets or multisets, but into equal length $k$-permutations (aka $k$-tuples).  The latter topic has been discussed here and is fairly standard in combinatorial terminology.
Lacking a more specific reference I will point out a pair of book length (or longer) treatments.  One of these is also mentioned in previous posts:
Stanley, Richard P. Enumerative Combinatorics, (Cambridge University Press, 1997/1999): two volumes
MacMahon, Percy A. Combinatory Analysis, (AMS Chelsea Publishing, 2001): two volumes bound as one book
The latter is a reprint of volumes by MacMahon originally published in 1915 and 1916.  This material was also covered in a 1978 work edited by George Andrews:
Andrews, George E. (editor) Percy Alexander MacMahon: Collected papers, Volume I, Combinatorics (Mathematicians of Our Time Series, The MIT Press, Cambridge, Massachusetts, 1978)
Andrews' book seems now out of print, but I suspect it is a very desirable way to appreciate MacMahon's work, esp. on partitions of multisets.  A 1980 review by John Riordan in Bull. AMS is available online by Project Euclid.  The Internet Archive also has images of MacMahon's original volumes.
