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

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?

The answer of $8$ seems to treat wearing distinguishable rings on two distinguishable fingers as equivalent to putting the rings in two distinguishable boxes, which is debatable. Arguably the answer should instead be $4!\cdot 2^3 = 192$ (taking into account order and orientation). If there's more to this question that you didn't show, it would be nice to give the whole statement. — Brian Moehring, Jun 01 '21 at 15:59
@rdev As I mentioned it comes from the order the rings are placed on the fingers (the $4!$ part) and the orientation of each ring (the $2^3$ part). Order would say $AB$ on a single finger is different from $BA$. Orientation would say taking off the ring and putting it back on the other direction is different (i.e. draw an arrow on top of the ring, is it pointing left or right?) — Brian Moehring, Jun 01 '21 at 16:23

Ritam_Dasgupta · Accepted Answer · 2021-06-01T16:17:55.083

2

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.

edited Jun 01 '21 at 16:17

answered Jun 01 '21 at 15:42

Ritam_Dasgupta

5,992
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I know, that's how the book explained it but the question doesn't say each finger must have a ring, hence the confusion. – rdev Jun 01 '21 at 15:51
Each finger may not have a ring, but each ring must be placed on a finger. I have written that in my answer. – Ritam_Dasgupta Jun 01 '21 at 15:52
why though? all the rings can be held in one finger too, no ?! – rdev Jun 01 '21 at 15:54
Please read my statement again, $\mathbf {carefully}$. – Ritam_Dasgupta Jun 01 '21 at 16:12
ya, got it. If you don't consider the combinations where you leave out a ring the answer does come out as 8. Thanks though ; ) – rdev Jun 01 '21 at 16:18
Yes, that is correct. – Ritam_Dasgupta Jun 01 '21 at 16:19

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?

1 Answers1