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I've been reading about Combinations, Variations and Permutations, all with and without repetitions. None of those seems to be what i'm looking for.

Given a set of 3 elements $(A, B, C)$, there are 6 possible Variations:

$1.\ (A,B,C)\\ 2.\ (A,C,B)\\ 3.\ (B,A,C)\\ 4.\ (B,C,A)\\ 5.\ (C,A,B)\\ 6.\ (C,B,A)$

I'm looking for all possible sequences without repetitions so $(A,B,C)$, $(B,C,A)$ and $(C,A,B)$ are all the same to me and i would simply write any (but only one) of them. The same happens with the other three, leading to only 2 possible sequences solutions: $(A,B,C)$ and $(A,C,B)$.

By a simple understanding, i know that the number of sequences without repetitions in a set of $n$ grouping by $r$ are: $\frac{1}{r} \cdot \frac{n!}{(n-r)!}$ because it's the number of Variations divided by the number of elements i'm grouping by.

EDIT: I found someone explaining this as "oriented cycles" here and also i've seen something similar defined as "cyclic permutations" on wikipedia.

Does this operation/function have a name?

LucianoSaldivia
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    I don't think your question is about groups in the sense of "group theory" (a group is a set together with a binary operation satisfying certain properties). You seem to be using "group" to mean "set". If so, the tags group-theory and graph-theory are both out of place. – Arturo Magidin Apr 28 '24 at 23:56
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    So $(A,B,C)$ is the same as $(B,C,A)$, but would not be the same as $(A,C,B)$. Is that correct? – Arturo Magidin Apr 28 '24 at 23:58
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    Are you asking: "how many permutations are there of $n$ elements if we consider permutations that can be obtained by cycling the elements to be the same?" This is $(n - 1)!$. I don't really understand what $r$ is in your post. – Izaak van Dongen Apr 28 '24 at 23:59
  • @ArturoMagidin Thanks for the clarification, i didn't know that and i was a bit lost. Yes, (A,B,C) and (A,C,B) are not the same, and having a set of 3 (grouping by 3) could only result in those 2 possible sequences. – LucianoSaldivia Apr 29 '24 at 00:40
  • @IzaakvanDongen In the example i gave, $n=3$ and $r=3$ as well so in this case both equations would give the same solution. $r$ is the number of elements in the resulting sequences. You could for example search in that same set, grouping by $r=2$, giving the results $(A,B)$, $(B,C)$ and $(C,A)$. Bear in mind that in this case Combinatorics and what i'm looking for are the same, but when $r \ge 2$ they differ as shown in the OP example with $r=3$. – LucianoSaldivia Apr 29 '24 at 00:48
  • @ArturoMagidin i just edited the post so that it's much more clear about the possible solutions. – LucianoSaldivia Apr 29 '24 at 00:57
  • You could call it the number of necklaces with a given set and number of beads (with the understanding that turning over a necklace makes it a different necklaces). Or, the number of ways of seating $r$ of $n$ different people around a circular table. – Gerry Myerson Apr 29 '24 at 05:21
  • The numbers are tabulated at https://oeis.org/A111492 with some discussion. – Gerry Myerson Apr 29 '24 at 05:24
  • @GerryMyerson That name makes a lot of sense, so i will be calling it that at least when explaining what i'm doing. I was actually looking for more of an 'official name' such as "combinations" or "permutations" but for this more specific case i guess it doesn't exist. Thanks for the helpful comment Gerry! – LucianoSaldivia Apr 30 '24 at 16:48
  • "Necklace" is actually a commonly used term in Combinatorics. https://en.wikipedia.org/wiki/Necklace_(combinatorics) and https://mathworld.wolfram.com/Necklace.html and https://www.britannica.com/science/necklace-problem and https://math.stackexchange.com/questions/1015418/number-of-necklaces-of-beads-in-two-colors and https://math.stackexchange.com/questions/314159/how-many-different-necklaces-problem and https://doc.sagemath.org/html/en/reference/combinat/sage/combinat/necklace.html and https://renrenthehamster.wordpress.com/2018/05/26/necklaces-and-bracelets-polya-enumeration-theorem/ – Gerry Myerson Apr 30 '24 at 22:22

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