Problem: Let $ A $ be a set of multisets of the same cardinality. Does there exist a bijection between $ A $ and the set of ordered pairs from all multisets in $ A $?
Examples:
$ A = \{ [1,1] , [1,2] , [2,2] \} $ $ \iff$ $ \{ \langle 1,1\rangle , \langle1,2\rangle , \langle2,2\rangle \}$ ( Here, $ A $ is a set of multisets generated from $ \{ 1,2 \} $ and all have cardinality 2)
$ A =\{ [1,1,1] , [1,1,2] , [1,2,2],[2,2,2],[1,1,3],[1,3,3],[3,3,3],[2,3,3],[2,2,3],[1,2,3] \} $
$ \iff$ $ \{ \langle1,1\rangle , \langle1,2\rangle , \langle1,3\rangle,\langle2,1\rangle,\langle3,1\rangle,\langle2,2\rangle,\langle3,3\rangle,\langle2,3\rangle,\langle3,2\rangle,\langle3,1\rangle \}$. ( Here, $ A $ is a set of multisets generated from $ \{ 1,2,3 \} $ and all have cardinality 3)
I came up with this question as I was working out combinatorics problems with multisets. Nevertheless , I'm not sure if the theory is correct ( and if so, how to formulate it ) and if not, do you know of a counter-example?
I think the answer is yes because any set of multisets of same cardinality is a representation of a stars and bars problem, but I haven't seen a concrete answer.
Let me try to reformulate what you appear to be asking. Given a finite set $S$ (in your examples $S=\{1,2\}$ respectively $S=\{1,2,3\}$) with $n$ elements, do the following two sets have the same cardinality: (1) the set of multisets on $S$ that with cardinality $n$, and (2) the set of ordered pairs of elements of $S$.
In your examples you find for $n=2$ the cardinalities $3$ for (1) and also for (2) (but that clearly is wrong since you forgot the ordered pair $\langle2,1\rangle$), and for $n=3$ you found $10$ for (1) and also for (2) (but that clearly is wrong since you wrote the ordered pair $\langle3,1\rangle$ twice). The general formulas are $\binom{2n-1}n$ for (1), and $n^2$ for (2), very different sequences of numbers. So the answer is no.