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The following combinatorial problem came up that seems easy but has turned out to be very intricate.

Assume that you are given $2n$ sets, say $S_1, \ldots, S_{2n}$ such that

  • each $S_i$ has exactly $n$ elements,
  • each element is in at most $n$ sets, $\forall x \colon | \{ S_i | x \in S_i \}| \leq n$,
  • for each $S_i$, there is an $S_j$ such that they are disjoint, $S_i \cap S_j = \emptyset$.

The final goal is to prove that it is possible to pick in each $S_i$ a different element $x_i \in S_i$ (i.e. $i \neq j \Rightarrow x_i \neq x_j$).

Assume we could prove that there is always a pair of elements $x,y$ such that none of the $S_i$ contains both $x$ and $y$. Then the statement above follows easily by induction.

But how to prove this? Is it actually true?

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Okay, the original problem can be solved using the marriage theorem. Using the marriage theorem, it is sufficient to prove that each collection of $S_i$ of size $k$ contains at least $k$ many different elements.

For $k \leq n$, there is nothing to do since already one single $S_i$ contains $n$ elements. For an arbitrary $k$, note that a collection of $k$ many $S_i$ contains $n \cdot k$ elements (counted with multiplicity). As each element occurs in at most $n$ many $S_i$, this means we have at least $\frac{n \cdot k}{n} = k$ many different elements.

The problem whether it is possible to pick a pair of elements such that none of the $S_i$ contains both remains unanswered.

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