TITLE: In how many ways $3$ rings can be worn in $2$ fingers if a finger can have any number of rings or no rings at all?
QUESTION [1 upvotes]: It has been calculated to be $8$, but I don't see how. If the rings are $A, B$ and $C$ then:
$\{(A,B), (A,C), (B,A), (B,C), (C,A), (C,B)\}$, that's $6$ ways, then if one finger has all the rings that's $2$ ways $\{ABC,0\}$ and $\{0,ABC\}$, so total $8$.
But what about the combinations taking two rings on one finger at a time: $\{(AB,C), (AC,B), (BC,A), (C,AB), (B,AC), (A,BC)\}$. That's another $6$ ways. Can someone please explain?

REPLY [2 votes]: I do not understand how you arrived at the first calculation. There you have taken the different possible duplets, but according to the question, each of the rings must be placed on one of the fingers. So that calculation was your mistake, as you haven't considered the position of the third ring.
The easiest way to solve these kinds of problems is this: Every ring can go to each of the two fingers, that is, there are $2$ possible choices for each ring. Since there are $3$ rings, total number of combinations is $2^3=8$.
Note: I have not considered the order in which the rings have been placed on the fingers, as @Brian Moehring says in the comments.
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