For any $k$-element multiset of $[n]$,
Let $a_i=\#\text{of occurrences of}~i~\text{in the multiset}$
We have then $a_1+a_2+\dots+a_n = k$ and each $a_i$ is a non-negative integer.
The set of solutions to the above equation are in direct bijection with the $k$-element multisets of $[n]$ using an obvious bijection: $(a_1,a_2,\dots,a_n) \leftrightarrow \{a_1\cdot 1,a_2\cdot 2,\dots,a_n\cdot n\}$ using multiplicity notation for multisets.
Now, the non-negative integer solutions to $a_1+a_2+\dots+a_n=k$ can be seen via "stars and bars", relating such a solution as a sequence of $n-1$ bars and $k$ stars. The bijection being
- $a_1=\#$of stars to the left of the first bar
- $a_n=\#$ of stars to the right of the final bar
- $a_i = \#$ of stars between the $i^{th}$ and $(i -1)^{st}$ bar for each other $i~$
So, the question has been reduced to how many sequences of length $n-1+k$ have exactly $n-1$ $B$'s and $k$ $S$'s. The answer being $\binom{n-1+k}{n-1}=\binom{n-1+k}{k}$
Generalizing the problem to allow the size of the multiset to change as well as the number of available numbers to use in the multiset.
As was started by another user, if we were to ignore the requirement on having at least one number occur exactly two times, we have the count being $\binom{n+k-1}{k}$
We wish to remove the "bad" multisets, which are those that do not have an element repeated exactly two times. To count how many are "bad" we can easily approach via generating functions. For each of the $n$ available numbers, the term $(1+x+x^3+x^4+x^5+\dots)$ will represent the available choices of taking zero, one, three, four, five, etc... copies of that number. With every number appearing any number of times except for two, there will clearly then be no number that occurs exactly two times.
The coefficient of $x^k$ then in the expansion of
$$(1+x+x^3+x^4+x^5+\dots)^n$$
will represent the number of multisets of size $k$ can be made using elements from the $n$ element set where none of the terms appear exactly two times.
The number of multisets which satisfy your condition will be $\binom{n+k-1}{k}$ minus the coefficient of $x^k$ in the series expansion of $(1+x+\frac{x^3}{1-x})^n$, or alternately worded using generating functions for the first, the overall generating function is:
$$\frac{1}{(1-x)^n}-(1+x+\frac{x^3}{1-x})^n$$
In your specific case we have for $n=3$ the series expansion
$$3x^2+6x^3+\color{red}{6x^4}+9x^5+13x^6+15x^7+18x^8+21x^9+\dots$$
The $6$ in $\color{red}{6x^4}$ corresponds to the six multisets you wrote down above.
Best Answer
The generating function for number $f(K,N,M)$ of multisets of cardinality $K$ from $\{1,\ldots,N\}$ with each element appearing at most $M$ times is
$$ g_{N,M}(z) = \sum_{K=0}^\infty f(K,N,M) z^K = (1+z+\ldots+z^M)^N = \left(\frac{1-z^{M+1}}{1-z}\right)^N$$
Now $$(1-z^{M+1})^N = \sum_{n=0}^N (-1)^n {N \choose n} z^{(M+1)n}$$ while $$(1-z)^{-N} = \sum_{k=0}^\infty {N +k-1 \choose N-1} z^k$$ so we get $$ f(K,N,M) = \sum_{n=0}^{\min(N,\lfloor K/(M+1) \rfloor)} (-1)^n {N \choose n} {N+K-(M+1)n-1 \choose N-1}$$