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I have a set $S$ of size $n=|S|$. I want the set of all $n!$ permutations of all $n$ elements of $S$, where permutations are represented as sequences, and not as bijections.

I have already seen this post, but it is not what I want. I don't want the permutation group or the symmetric group, I simply want the set of sequences.

For instance, $S=\{1,2,3\}$, and I want to obtain $\{(1,2,3),(1,3,2),(2,1,3),(2,3,1),(3,1,2),(3,2,1)\}$.

Is there a notation to do that, and if not, is there a formula to obtain what I want?

Thank you for your help.

JacopoStanchi
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  • A "formula" to obtain what? If you mean the number of those permutation then it is easy to calculate, but if you mean the list of them, a formula is not what you are looking for. Formulas usually output values, not lists. – Zima Jun 21 '23 at 14:50
  • Not the number (I know it is $n!$), but the set of all permutations as I said. I assume that there should be some sort of formula, because a naive solution could be to do the set of all possible sequences of size $n$ using elements of $S$ (that is, $S^n$) and then remove all sequences with repetitions. – JacopoStanchi Jun 21 '23 at 14:53
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    https://en.wikipedia.org/wiki/Steinhaus%E2%80%93Johnson%E2%80%93Trotter_algorithm – Mike Earnest Jun 21 '23 at 14:55
  • @MikeEarnest Thank you, but I'm not really looking for an algorithm. I'm looking for a conventional notation (for instance $\mathit{Permutations}(S)$) if there exists one, or a formula otherwise. – JacopoStanchi Jun 21 '23 at 15:01
  • @JacopoStanchi no formula could outbut the list you want – Zima Jun 21 '23 at 15:02
  • So, this is just a language/notation question? If so, who is your audience, or what is the context in which you need to describe this list? Please answer by editing the question body, not in a comment. – Mike Earnest Jun 21 '23 at 15:03
  • Thank you for all your answers, but I thought that by writing "Notation" in the title, by adding the "notation" tag, and by saying "Is there a notation to do that" in the question body it would be clear enough... – JacopoStanchi Jun 21 '23 at 15:05
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    $S_n$ is the symmetric group on $n$ elements which consists of the $n!$ permutations of those $n$ elements. You can choose to interpret the elements of $S_n$ however you like... as bijections, as sequences, as candy canes... however you like. We in mathematics freely switch between interpretations, just like how we might think of "the number $1$" as the natural number $1$, the integer number $1$, the rational number $1$, and so on... depending on the context we are in... all versions of which having some sense of being "the number 1" – JMoravitz Jun 21 '23 at 15:11
  • While to some people there may be a technical difference between the function $f~:~{1,2,3}\to{1,2,3}$ where $n\mapsto n$ and the sequence $(1,2,3)$... in effect there is not enough of a difference to care about there being a difference between these objects... and in actuality, rigorously since the way sequences are defined in the first place, there is not even a rigorous distinction between these objects... they happen to be identical. – JMoravitz Jun 21 '23 at 15:14
  • So... even though you say in your post "I don't want the permutation group or the symmetric group" the real answer is that you do want those and you just did not realize it, having thought that there was some esoteric distinction when there was none. – JMoravitz Jun 21 '23 at 15:17

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