Suppose we have a $n$-element set $S$. Denote the set of its $k$-element subsets by $K$ ($|K|=\binom{n}{k}$).

If the elements of $S$ are real numbers then to each $k$-element subset we can associate its sum. If these sums $\binom{n}{k}$ numbers are all distinct, they induce an order on $K$.

What is number of orders on $K$ we can obtain this way?

It is clear that this is not all $\binom{n}{k}!$ orders are possible. For example, because the smallest $k+1$ elements defined by the order of numbers of $S$ and $n! < \binom {\binom{n}{k}}{k}$ for sufficient large $k$. There are many other constrains on the orders of $K$.

For $n=4$, and $k=2$. $S=\{a, b, c, d\}$ and $K=\{(a,b),(a,c),(a,d),(b,c),(b,d),(c,d)\}$. If $\{a, b, c, d\}=\{0.2, 0.4, 0.8, 0.3\}$, then the correspondent sums equals $\{0.6,1.0,0.5,1.2,0.7,1.1\}$, and we have the following order on $K$: $\{(a,d),(a,b),(b,d),(a,c),(c,d),(b,c)\}$.

Note that we can not get the order $\{(a,b),(a,c),(c,d),(b,d),(a,d),(b,c)\}$ because from the order of the first two elements we get that $b<c$, and from the second two element we get $b>c$.