What is the generating function for a bin that has either zero elements or 2 only?
We start with: $(1+x^2)$ which if it had an $x$ it would translate to $\frac {1-x^3}{1-x}$ So I thought maybe I could represent $(1+x^2)$ like so: $(1+x^2)=(1+x+x^2)-((1+x)-(1))$. I know what first two translate to but what $(1)$ translate to? (It's an empty bin) is it: $\frac {1-x}{1-x}=1$?
So the generating function for $(1+x^2)$ would be: $\frac {1-x^3}{1-x} - \frac {1-x^2}{1-x} +1$?
If your question was counting the number of ways of distributing $n$ indistinguishable balls among $m$ distinguishable bins so each bin had $0$ or $2$ balls, the answer would be the coefficient of $x^n$ in the expansion of $(1+x^2)^m$.
If you let $m=1$, you are left with looking at one bin and the generating function $(1+x^2)$.
There is one way of distributing $0$ indistinguishable balls so the bin has $0$ or $2$ balls (the coefficient of $x^0$) and one way of distributing $2$ indistinguishable balls so the bin has $0$ or $2$ balls (the coefficient of $x^2$), and no ways of distributing any other number of balls so the bin has $0$ or $2$ balls. It does not get any more complicated than that, and you do not need to complicate $(1+x^2)$.