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In Stanley's Enumerative Combinatorics, he defines a permutation $w$ of the set $S=\{x_1,...x_n\}$ with cardinality $n$ to be linear ordering $w_1w_2...w_n$, so that the word $w=w_1w_2...w_n$ corresponds to the bijection $$w:S\to [n]=\{1,2,...,n\},\quad w(x_i)=w_i$$ As an example, one permutation of $S=[4]$ is the word $4123$, which corresponds to the mapping $w(1)=4,w(2)=1,w(3)=4,w(1)=2$.

Extending this to permutations of multisets, Stanley says that we again think of $w$ as a word $w_1w_2...w_n$, where $n$ is the cardinality of multiset $M$.

I have no issue with this definition of permutations of multisets. This question is about extending to multisets the association of a permutation with a unique bijection; how do we do it?

This post seems to suggest that we may do so by defining it as a bijection from subsets of $M$ to subsets of $[n]$, i.e. we associate $w=1213$ with the mapping $w(\{1,1\})=\{1,3\},w(\{2\})=\{2\},w(\{3\})=\{4\}$. Is this correct?

Edit: These was some discussion of forming equivalence classes of bijections in a now-deleted answer. When describing multisets, Stanley writes that for a permutation of $M=\{x_1^{a_1},\ldots,x_m^{a_m}\}$, where $a_i$ denotes the multiplicity,

if $x_i$ appears in position $j$ of the permutation, then we put the element $j$ of $[n]$ into Category $i$;

this was done to enumerate the permutations (the multinomial). Stanley does not discuss forming equivalence classes between bijections; if this is what he is doing, it it unclear to me what the equivalence relation is, and how it is an equivalence class of similar bijections.

David Raveh
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  • Is there any specific reason that you want to regard $w$ as a bijection instead of a mapping without uniqueness? – Vezen BU Jul 14 '23 at 07:47
  • @VezenBU well, we need to associate every permutation with a single mapping so the number of mappings is the multinomial. Unless I define an equivalence class...? Is this what you mean? – David Raveh Jul 14 '23 at 12:03

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