# Counting multisets satisfying a fixed property

Suppose $$S$$ is a infinite set and $$R\subset S$$ is also infinite. Now, we want to find the number of multisets $$(M,\nu)$$, with $$M\subset S, |(M,\nu)|=n$$, and having an additional property that for every $$x\in M$$ with $$2\nmid \nu(x)$$, we must have, $$x\in R$$. How can we find the number of all such multisets $$(M,\nu)$$ of cardinality $$n$$. I have no clue how to tackle this. I am looking for a generating function for $$f(n)$$, where $$f(n)$$ denotes the required number for every $$n\in \mathbb{N}$$.

Suppose, if we omit the additional property, that is, if we try to find all multisets $$(M,\nu)$$ with cardinality $$n$$, then it is simply the coefficient of $$u^n$$ in

$$\prod_{u\in S} (1+u+u^2+\ldots)=\prod_{u\in S}(1-u)^{-1}$$ I don't understand how to get my result, by using the above, or otherwise.

Note: A multiset is a pair $$(M,\nu)$$, where $$M$$ is a set, and $$\nu:M\to \mathbb{N}$$. By cardinality of $$(M,\nu)$$, we mean $$\sum_{x\in M}\nu(x)$$.

Thank you for any kind of help!

In other words, every element $$x\in M\cap (S\setminus R)$$ has even multiplicity $$\nu(x)$$, while the multiplicity of elements of $$M\cap R$$ is unrestricted.
Then, the generating functions is $$\begin{split} \sum_{n\geq 0} f(n) x^n &= \prod_{u\in S\setminus R} (1+x^2+x^4+\dots)\prod_{u\in R} (1+x+x^2+\dots) \\ &= \prod_{u\in S\setminus R} (1-x^2)^{-1} \prod_{u\in R} (1-x)^{-1} \\ & = (1-x^2)^{-|S\setminus R|}(1-x)^{-|R|} \\ &= (1+x)^{-|S\setminus R|}(1-x)^{-|S|}. \end{split}$$
This, however, makes sense only for finine sets $$S$$ and $$R$$, since if they are infinite, so is $$f(n)$$.