Let $r\_1, r\_2, \ldots, r\_k$ be positive integers with or without repetition such that $1\le r\_i \le n$ for $i = 1, 2, \ldots, k$. Let $f$ be a continuous multivariate function with the property that the value of $f(r\_1, r\_2, \ldots, r\_k)$ is independent of the order the arguments $r\_1, r\_2, \ldots, r\_k$. The trivial examples are:


$f(r\_1, r\_2, \ldots, r\_k) = g(r\_1) + g(r\_2) + \ldots + g(r\_k)$,


$f(r\_1, r\_2, \ldots, r\_k) = g(r\_1)g( r\_2) \ldots g(r\_k)$ and


$f(r\_1, r\_2, \ldots, r\_k) = c$


where $g$ is a continuous univariate function and $c$ is a constant. I have two questions


**Questions:**


1. Are there other non trivial examples of such functions?
2. Are there infinitely many such non trivial functions $f$?



Multivariate functions whose value is independent of the order of the arguments
You could symmetrize an arbitrary function by averaging it over the symmetric group.


Explicitly: take your favourite function $h$ in $n$ variables, and set $f(r\_1, ..., r\_n) = \frac{1}{n!}\sum\_{\sigma} h(r\_{\sigma(1)}, ..., r\_{\sigma(n)})$ where the sum is taken over all permutations $\sigma$ of the numbers $1,...,n$. 


I'd love to hear of an example which *isn't* obtained by symmetrization! 


edit: heh. Any such example would *be* its own symmetrization. So I guess in some sense this is a complete, if overly glib, answer.


