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Count of permutation of n different items are:

A(n) = 1*2* .. *n

e.g:

  • [1] -> {1} // 1
  • [1,2] -> {1,2}, {2,1} // 2
  • [1,2,3] -> {1,2,3}, {1,3,2}, {2,1,3}, {2,3,1}, {3,1,2}, {3,2,1} // 6

So what is the permutation of: [], (aka. empty input).

I've checked this question: Does the set of permutations of an empty set contain an empty set?
The answer said it should be 1, (aka. {}), and not 0, but what's the proof?

Eric
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    Does this answer your question? Does the set of permutations of an empty set contain an empty set? – Another User Mar 29 '23 at 08:34
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    To understand the answer to the question you reference you have to understand what it means to be a map from a set $X$ to a set $Y$: it is a subset of $X\times Y$ that satisfies a certain property. A permutation then is just a $1--1$ onto map from $\emptyset$ to $\emptyset$, and $\emptyset\times\emptyset=\emptyset$, so the only possible map is the set $\emptyset$. You then have to check it is (i) a map (ii) 1--1 (iii) onto. The details are in elementary set theory textbooks. – ancient mathematician Mar 29 '23 at 08:57
  • @AnotherUser No, I already mentioned that post in my question. – Eric Mar 29 '23 at 09:27
  • @ancientmathematician What u said make sense to me in some degree, because in the iterative solution in my golang code, there need a start point which is not empty. Refer: https://stackoverflow.com/a/75875105 , but still I want some kinda of more sufficient proof. – Eric Mar 29 '23 at 09:29
  • @Eric I have answered your maths question, I have no idea about coding. Your last sentence is not clear. The set of permutations of the empty set has precisely one element $\emptyset$, it is the set ${\emptyset}$, aka $1$. You seem to think that $1$ is ${}=\emptyset$ (this set is $0$). – ancient mathematician Mar 29 '23 at 11:28
  • No, I know the 1 means {∅}, just as I initialized the list in code combList := [][]T{{}}, thanks anyway, I would check more about elementary set theory. – Eric Mar 29 '23 at 12:39

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