# Number of ways to put an indefinite number of balls of two colors into $n$ bins

I believe it's likely this question has been asked before, but I lack the skill to search for it. Suppose I have an indefinite supply of red and green balls. I have $n$ bins, and I want to know the number of ways I can place either 0 balls, one red, one green, or both red and green into each bin (each bin will be in one of these four states). Order does matter. I know this involves using the binomial theorem, but I don't know how to account for the fact that each bin may contain at most one of each color ball. I feel like the number of balls doesn't actually matter, since it's "enough" to maximally fill each bin.

Consider the $2n$ symbols

$$1_{\text{red}}, \dots, n_{\text{red}}, 1_{\text{green}}, \dots, n_{\text{green}}$$

where $i_{\text{red}}$ or $i_{\text{green}}$ means place a red ball or a green ball into bin $i$, respectively.

If order matters, and we are only allowed to put at most one red and one green ball into any given bin, then we are just taking a string of these $2n$ symbols without duplicates.

There are $$\frac{(2n)!}{k!}$$ strings (permutations) of length $k$ so the total is

$$\sum_{k = 0}^{2n} \frac{(2n)!}{k!}.$$

This is the even index part of sequence A000522 in the OEIS if you want to look for more information such as alternative formulae. For instance,

$$\sum_{k = 0}^{2n} \frac{(2n)!}{k!} = \lfloor (2n)! e \rfloor$$

for $n \ge 1$.