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PrintInternational Mathematical Olympiad
China counting and probability
Problem
Let be a subset of the set containing exactly elements. Prove that there exist numbers in such that the sets are pairwise disjoint.
Solution
Consider the set . There are at most elements in . Two sets and have nonempty intersection if and only if is in . So we need to choose the elements in such a way that the difference for any two elements is not in .
Now select these elements by induction. Choose one element arbitrarily from . Assume that elements, , are already chosen. An element in that is already chosen prevents us from selecting any element from the set , where . Thus after elements are chosen, at most elements are forbidden. Since has elements, and , there is always at least one element left to choose for the next . Therefore, we can choose such elements so that the sets are pairwise disjoint.
Now select these elements by induction. Choose one element arbitrarily from . Assume that elements, , are already chosen. An element in that is already chosen prevents us from selecting any element from the set , where . Thus after elements are chosen, at most elements are forbidden. Since has elements, and , there is always at least one element left to choose for the next . Therefore, we can choose such elements so that the sets are pairwise disjoint.
Techniques
Games / greedy algorithmsInduction / smoothingColoring schemes, extremal arguments