Number of ways to partition a multiset into $k$ non empty submultisets. Let $A$ be a multiset with $n$ distinct elements where each element occurs exactly twice. How many ways can we partition $A$ into $k$ non-empty (unlabelled) sub multisets (denoted $T(n,k)$)?
My approach would be something similar to the Stirling Numbers. For each element $x \in A$, we can either both copies of $x$ in a set, or include them in two different sets. Therefore, we can define labelings of sets (parts), as singleton and two-element subsets of $\{1,2,3,...k\}$. There are in total $k + {{k}\choose{2}} = {{k+1}\choose{2}}$ different labeling we can assign to each element. (Note: We divide the end result by $k!$ because the labeling did not originally matter) Elements where both copies are included in the same set are labeled with a singleton, and if one element occurs in two sets, it is labeled with a two-element set.
For example, the partition of $\{\{a,a,c\},\{b,b,c\}\}$ of $\{a,a,b,b,c,c\}$ can be defined by an equivalence class or function such as $f(a)=\{1\}$, $f(b)=\{2\}$, $f(c)=\{1,2\}$.
The basic idea is to count the number of functions $f:A_s \xrightarrow{} S$ such that $|\cup_{x \in A_s} f(x)| = k$. Here, $A_s$ is the set containing only one of each element in $A$, and $S = \{  s \in \mathcal{P}(\{1,2,3,...k\})\ \mid  |s| = {1,2} \}$.
Since $|S|={{k+1}\choose{2}}$, and $|A_s|=n$, we have $|S|^{|A_s|} = {{k+1}\choose{2}}^{n}$ different functions to choose from.
However, some functions may not satisfy our initial constraint that $|\cup_{x \in A_s} f(x)| = k$. We can use inclusion-exclusion for this (similar to how the Stirling Numbers of the Second Kind are derived).
What I get is something like

$$ T(n,k) = \frac{1}{k!} \sum_{i=0}^{k} (-1)^{k-i} {{k}\choose{i}} {{i+1}\choose{2}}^n $$

I think the formula is wrong though, and I can't figure out why.
For example, with $k=2$, we have $T(n,2) = \frac{1}{2}(3^n - 2)$. I know that it should be $\frac{1}{2}(3^n - 1)$ because each subset of $A$ has a complement, but one set is its own complement, and we need to "add" a pair to the collection, then divide by 2.
For $k=3$, we have $T(n,k) = \frac{1}{6}(6^n - 3^{n+1} + 3)$, but manual computations show that this is incorrect. (For example, $T(3,3) = 23$, but it should be $26$, and $T(4,3) = 176$, but it should be $183$).
Could anyone please kindly give me hints on what I am missing here. I am really trying to figure it out on my own, or at least understand why my computations are incorrect? Thanks in advance!
Edit:
I realized my mistake (thanks to user2661923).
So basically, I had the counting correct, but for ordered partitions rather than unordered partitions. Basically, if we forget the ordering, some partitions have $k!$ copies, others have fewer than $k!$ duplicates.  So all we need to do to fix the counting is "add" enough duplicates (for those permutations that have fewer than $k!$ copies).
For example, if $k=2$ using the formula above (ignore the $\frac{1}{k!}$), we get $T(n,k) = 3^n-2$. However, there will be one partition of the form $\{X, X\}$ (where $X$ is a multiset). There is only one copy of this partition included, so we need to "add" in another copy of it, leading to $3^n-1$, instead of $3^n-2$, which we can then divide by 2, to get the correct value of $T(n,2) = (3^n-1)/2$
If $k=3$, then we initally get $T(n,3) = 6^n - 3^{n+1} + 3$. Partitions of the form $\{X,X,Y\}$ only have $3$ copies included, so we need to "add" three more. Each element is either assigned to $\{X,X\}$ or $\{Y\}$, but all cannot be assigned the same sets, so this yields $2^n-2$ different copies we need to add (don't forget that this gives $3$ additional copies). Since we originally had $3$ different labels, we need to multiply this result by $3$. This gives $$T(n,3) = (6^n - 3^{n+1} + 3 + 3(2^n-2))/6 = (6^n - 3^{n+1} + (3)2^n - 3)/6$$
The case for $k=4$ is more complicated, but after manually checking all possible partitions which need extra copies, I get the formula:
$$T(n,4) = (10^n - (4)6^n + (6)4^n - (9)2^n + 8)/24$$
I'm not putting this as an answer because I am still not sure of a general formula that doesn't involve manually checking all possible partitions which are initially undercounted.
 A: I have finally reached the point where I have a clear understanding of the intent of the OP (i.e. original poster), and can therefore identify his fatal error.  His error is in assuming that
$$\frac{\text{Satisfying distributions when multisubsets are labeled}}{\text{Satisfying distributions when multisubsets are not labeled}} = k!. \tag1 $$
The assertion in (1) above is flat wrong.  For illustration purposes, I will assume that $~n = 5,~$ and 
$~A = \{x_1,x_1,x_2,x_2,x_3,x_3,x_4,x_4,x_5,x_5\}.$
I will also assume that when the multisubsets are labeled, the labels are $B_1, B_2, \cdots, B_k.$

$\underline{\text{Example 1}}$
Assume that $k = 2$, and consider the distribution of

*

*$B_1 = \{x_1,x_2,x_3,x_4,x_5\}.$

*$B_2 = \{x_1,x_2,x_3,x_4,x_5\}.$
Here, $B_1,B_2$ are identical to each other.  So, with this specific distribution, the ratio referred to in (1) above is in fact $\displaystyle \frac{2!}{2!} = 1$.
That is, since $B_1$ and $B_2$ are identical, permuting the labels of the submultisets, so that the first group of elements goes in $B_2$ rather than $B_1$ results in the exact same distribution.

$\underline{\text{Example 2}}$
Assume that $k = 3$, and consider the distribution of

*

*$B_1 = \{x_1,x_1,x_2,x_2\}.$

*$B_2 = \{x_3,x_3,x_4,x_5\}.$

*$B_3 = \{x_4,x_5\}.$
Here, $B_1,B_2,B_3$ are all distinct.  So, with this specific distribution, the ratio referred to in (1) above is in fact $(3!)$.
That is, there are $(3)$ ways of determining which of the three submultisets will contain $\{x_1,x_1,x_2,x_2\}.$
Once this is done, there are then $(2)$ ways of determining which of the two remaining submultisets will contain $\{x_3,x_3,x_4,x_5\}.$

$\underline{\text{Example 3}}$
Assume that $k = 3$, and consider the distribution of

*

*$B_1 = \{x_1,x_1,x_2,x_2,x_3,x_3\}.$

*$B_2 = \{x_4,x_5\}.$

*$B_3 = \{x_4,x_5\}.$
Here, $B_2,B_3$ are identical to each other.  So, with this specific distribution, the ratio referred to in (1) above is in fact $\displaystyle \frac{3!}{2!} = 3$.
That is, there are $(3)$ ways of determining which of the three submultisets will contain $\{x_1,x_1,x_2,x_2,x_3,x_3\}.$
Once this is done, the two remaining submultisets will each contain $\{x_4,x_5\}.$
