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Austria 2010

Austria 2010 counting and probability

Problem

We are given the set of all non-negative integers less than or equal to . We call a subset of outstanding if it is not empty and a -element subset of exists for all . Determine the number of outstanding subsets of .
Solution
If is the largest element of an outstanding subset , it follows that must contain elements. This is possible if it either contains all elements not greater than or all but one. We see that each outstanding subset of corresponds to an ordered pair of integers with , whereby the case of an element subset with maximum element is denoted by the pair and an element subset with maximum element and missing the number is denoted by . We also note that each such pair corresponds directly to an outstanding subset. The number of outstanding subsets is therefore equal to the number of ordered pairs of this type. This number is equal to the number of 2-element subsets of plus the number of elements of . We see that the number of outstanding subsets is equal to
Final answer
binom(n+2, 2)

Techniques

Recursion, bijection