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.
The problem here is that $a$ and $c$ have two different roles. Therefore simply using $\binom{5}{2}$ will not work, as it denotes the number of ways of picking two elements out of five, but without distinguishing between them. What you can do is
$$\binom{5}{2}\cdot2\cdot4 = 80\ ,$$
where we piked two numbers in $\{1,2,3,4,5\}$ using $\binom{5}{2}$, then we choose which one to assign to $a$ and which one to $c$ (whence the factor $2$), and finally we choose where to put the $c$ (the factor $4$).
Best Answer
Revised in response to the clarification.
There are $\binom{n}2$ different pairs. Once we’ve chosen one of them, there are $\binom{n-2}2$ ways to choose a pair from the remaining $n-2$ balls, so there are $\binom{n}2\binom{n-2}2$ ordered pairs of boxes with disjoint contents. Since we want unordered pairs of boxes, this has to be divided by $2$: there are
$$\frac12\binom{n}2\binom{n-2}2=\frac{n(n-1)(n-2)(n-3)}{8}$$
possible pairs of boxes with disjoint contents.