# Whats wrong with this reasoning...

Suppose I have two non-distinguishable balls (for example two white ones) and I color them with red and green, then a combinatorial reasoning could go like this.

1. Suppose I enumerate the balls, ball one and ball two, and by this distinguish them, and there are $2^2 = 4$ ways to color them
2. Because they are indistinguishable we had to divide by the number of enumerations, which are $2!$

So in total we have $2^2 / 2! = 2$ different colorings. But obviously wrong, we have

red, red red, green green, green

as different coloring. So what is wrong with this reasoning, which as I see is frequently applied in combinatorial problems?

EDIT: Please consider my other post of a non-trivial conclusion drawn by such an argument, which confuses me cause the proof should be correct...

• When you divide "by the number of enumerations", you only had to divide the RG, GR colorings (by $2$). That is, these two colorings count as the same and thus overcount. The other two cases, $RR$ and $GG$, did not overcount. Mar 13 '14 at 19:44
• so in a general scheme or formula we should first count the elements with different entries, and divide them by #number of entries!? Mar 13 '14 at 19:56
• @Stefan: Yes, something like that. For more fun, look up Burnside's lemma and Polya counting sometime. It may help here to not thinking of "dividing" at all, but simply say how many times each colouring has been counted in your "ways". It turns out that $\{R, G\}$ has been counted twice (once as RG and once as GR), but $\{R, R\}$ and $\{G, G\}$ have each been counted only once. Mar 13 '14 at 20:19

• by pair you mean set, then there are $\binom{10}{2} = 45$ ways. Mar 13 '14 at 20:09
• then it is: $\binom{10}{2}\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2} = 113400$? Mar 13 '14 at 20:24