Distinguishability in Round Table Combinatorics I have stumbled upon many questions, and one of the weaknesses is the ability to test if the concept is distinguishable or not. For example this: 

Nine delegates, three each from three different countries, randomly select chairs at a round table that seats nine people. Let the probability that each delegate sits next to at least one delegate from another country be $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Take the race: $AAA$ Then are the people distinguishable or not? Is it: $A_1, A_2, A_3$ or not? I ask this because of cyclic shifts. I asked this question Here.. But drhab said they are not, but he uses the concept of $AaA$, or $AAa$ or $aAA$? Is there something with the chair being distinguishable or what is the whole issue exactly?
 A: I think this is an important question; there's a lot of confusion on this issue that comes up regularly in other questions, and it's good to have answers to it in one place here.
Two different cases should be distinguished: Determining probabilities and counting "ways" (of arranging/combining/doing/... things). As true blue anil has pointed out, when you want to determine probabilities, it doesn't matter what you treat as distinguishable. Treating things as distinguishable just multiplies the numerator and the denominator by the same factor when you divide the number of favourable cases by the total number of cases.
What does matter is which events you treat as fundamentally symmetric and hence equiprobable. For instance, if you flip two coins, there are $4$ possible ordered tuples ($(H,H)$, $(H,T)$, $(T,H)$ and $(T,T)$) but only $3$ possible unordered tuples ($(H,H)$, $(H,T)$ and $(T,T)$). This has nothing to do with whether you or anyone else can distinguish the coins. You can make the coins as indistinguishable as you like, you can have a machine flip them and then shuffle the flipped coins before you get to see them – none of this will change the fact that the four ordered tuples are equiprobable and the three unordered tuples are not, and you'll get wrong probabilities if you treat the unordered tuples as equiprobable.
That was all about probability; now to counting "ways". In that case, distinguishability is entirely a matter of definition. The question whether it is "correct" to treat things as distinguishable, when posed in this context, is meaningless. The term "distinguishable" is slightly unfortunate here, as it sounds as if it has something to do with our sensory abilities. A better term might be "distinct". The counting task must specify which "ways" are to be counted, and this must include a definition of which "ways" are to be regarded as distinct.
Perhaps it's worthwhile to add that in quantum theory indistinguishability comes into play for particles in a far more profound way than it ever does for coins. Quantum statistics can be derived by requiring that states are either completely symmetric (for bosons) or completely antisymmetric (for fermions) with respect to exchange of indistinguishable particles, with the result e.g. that there are only three and not four possible combined spin states of two bosons with two spin states each, namely $\uparrow\uparrow$, $\frac12(\uparrow\downarrow+\downarrow\uparrow)$ and $\downarrow\downarrow$, and they are equiprobable, leading to an increased probability for states in which the bosons share the same state. So in a sense bosons really do behave like indistinguishable coins would if their indistinguishability had anything to do with probabilities.
