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.
- 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
- 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...
The colorings are as follows:
(Red, Red), (Red, Green), (Green, Red), (Green, Green). As the balls are unordered, the colouring (Red, Green) and (Green, Red) are the same, and then you are overcounting.
Think of another similar problem: You have 10 different players, and you want to put them in five pairs. How many different pairs can you form?