Size of a Power Set of a Multiset where some items are Indistinguishible

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Note: I do not know all the terminology for my question, so I am making some educated guesses. Please let me know the correct terminology for anything I get the name of wrong.

Suppose I have a multiset, say, $ S = \{1, 1, 2, 3\}$. I want to find the size of the power set - and importantly, not the power multiset. Further, I want each item in the power set to be treated as a multiset.

Example of the power multiset (what I do not want):

$$\{\{\}, \{1\}, \{1\}, \{2\}, \{3\}, \{1, 1\}, \{1, 2\}, \{1, 3\}, \{1, 2\}, \{1, 3\}, \{2, 3\} \{1, 1, 2\}, \{1, 1, 3\}, \{1, 2, 3\}, \{1, 2, 3\} \{1, 1, 2, 3\}$$

and its size may be determined without calculating the power set itself with $2^{|S|}$ - here, 16.

However, what if we do not want duplicates, because the 1s are not distinguishable?

Example of two indistinguishable elements of the power set: $$\{1\}, \{1\}$$ - because the two sets are identical.

Example of two distinguishable elements of the power set: $$\{1, 1, 2\}, \{1, 2\}$$ - because, being multisets, the existence of two 1s is not the same as the existence of one 1.

In our example, the power multiset would be $$\{\{\}, \{1\}, \{2\}, \{3\}, \{1, 1\}, \{1, 2\}, \{1, 3\}, \{2, 3\}, \{1, 1, 2\}, \{1, 1, 3\}, \{1, 2, 3\}, \{1, 1, 2, 3\}\}$$

and its size is 12, but I don't know of any formula to figure that out ahead of time - I wrote it out and then counted the elements by hand.

Given a multiset, is there a formula to find the size of its power set without explicitly constructing it? This question is similar, but (1) it is asking about combinations of a specific size, not the whole power set and (2) embarrassingly, I couldn't understand the answer. This question is also asking the same thing, but only for pairs. (I suspect that a recursive formula exists.)

(For context: I'm trying to find the number of factors of large numbers, and determined that it is equal to the size of the power set of the multiset of its prime factorization, and now am hoping for an efficient way to find it.)