Ramsey's theorem for $c$ colours states that (this statement is not copied from anywhere):
Let $c\ge 2$ be an integer, $[c]=\{1,2,\ldots,c\}$ represent $c$ colours, and let $(r_1,r_2,\ldots,r_c)$ be a list of integers greater than or equal to $2$ (note: the integer $1$ can be allowed among the $r_i$ but I prefer to not include this degenerate case). For all sufficiently large positive integers $n$ and for any colouring of the complete graph $K_n$ by the colours $[c]$, there exists an index $i\in[c]$ such that this colouring of $K_n$ contains a monochromatic $i$-coloured $r_i$-clique.
The smallest such $n$ is denoted by $R(r_1,r_2,\ldots,r_c)$. Now suppose the colours stayed the same but we swapped some $r_i$ and $r_j$ in the list $(r_1,r_2,\ldots,r_c)$. Then the two Ramsey numbers stay the same by an argument involving inverse colourings. Since every permutation can be represented by a composition of transpositions, it does not matter how we order the $(r_1,r_2,\ldots,r_c)$. A couple of questions related to each other that are on my mind are:
- Okay... so the $r_i$ are independent of order. But if we put them into a multiset where repetition is allowed but order does not matter, then is each $r_i$ still related to a specific colour?
- Is it true that the colour just need to be distinct once we pass the $(r_1,r_2,\ldots,r_c)$ to a multiset, or does the order on the colours still matter?
These questions are inspired by the fact that when $r_1=r_2=\cdots=r_c$ have a common value $r$, then $R(r_1,r_2,\ldots,r_c)$ has to do with the existence of a monochromatic $r$-clique without caring about the particular colours; the only thing that matters is the size of the desired clique and the number of acceptable distinct colours.
Basically, what I'm wondering is the extent to which we can do some "modding out" in Ramsey's theorem after the usual quoted theorem is obtained, either with respect to the $r_i$ or the colours themselves, or both. I just can't wrap my head around what is the possible statement of the symmetry that seems to exist.