This is what I mean: I have three numbers $[a,b,c]$ and I can produce a set from the numbers comprised of pairs of $1$ number and the sum of the other two numbers, ie. $\{(a, b+c), (b, a+c), (c, a+b)\}$. With four numbers $[a,b,c,d]$, the set becomes $\{(a, b, c+d), (a, c, b+d), (a, d, b+c), (b, c, a+d), (b, d, a+c), (c, d, a+b)\}$.
Edit: The pairs in the set are unordered, so $(1,2)$ is the same as $(2,1)$
I need to be able to give a general definition of this set for any size of the initial set of numbers. This is for a proof that would involve that set for any size.
For some background, I am trying to prove that for three numbers $[a,b,c]$ no two pairs from the set can contain the same number. For four numbers $[a,b,c,d]$, no two triples from the set can contain two of the same numbers. For five numbers $[a,b,c,d,e]$ no two elements can contain three of the same numbers, etc. I have proved this for $3$ and $4$, but it gets very tedious for $5$. (I am not sure that it works for larger numbers, so if anyone knows that it doesn't (or if there is already a proof like this out there somewhere) please let me know and the question is moot.)
This is how I have started:
For any set of integers $I=(i_1, i_2, i_3, ..., i_n)$ used to generate a set of the form $\{(i_1,\dots, i_{n-2},(i_{n-1}+i_{n})),(i_1,\dots, i_{n-3}, i_{n-1}, (i_{n-2}+i_{n})),\dots)\}$
But I'm not sure how to define the full set, or ever if it can be defined in such a way. I can easily define it in words as "the set of (pairs, triples, etc.) containing all but two of the integers from $I$ and the sum of the remaining two", but this really doesn't help me with my proof at all.