Problem:
Find the number of four-element subsets of with the property that two distinct elements of the subset have a sum of , and two distinct elements of the subset have a sum of . For example, and are two such subsets.
Solution:
There are two types of that have the needed property. There is either an assignment of distinct values for , and such that and or an assignment such that and . These two types are mutually exclusive because and imply that . For the first type, there are choices for , namely , and , and there are choices for , namely , , and . Thus a four-element subset of the first type can be formed by taking the union of one of two-element subsets with one of two-element subsets as long as those two subsets are disjoint. There are such pairings that are not disjoint out of the pairings, so there are subsets of the first type.
For subsets of the second type, there are choices for a value of , ) such that and can be two other elements of the subset. Note that in each of these cases, . For each of these, there are other values that can be chosen for the element in the subset. But counts some subsets more than once. In particular, a subset is counted twice if or . In such cases either or . There are exactly subsets where the role of can be played by two different elements of the set. They are , and . Thus there are subsets of the second type.
In all, there are subsets with the required property.
The problems on this page are the property of the MAA's American Mathematics Competitions