Team Selection Test

The answer is $1^{2012}+2^{2012}+\cdots +19^{2012}$.

We first observe that the minimum element of $U\cup V$ is the minimum of the minimum elements of $U$ and $V$ for all finite sets $U$ and $V$ of integers.

Let the value of $f(A)$ be $n$. Then by the given property of $f$ we have $f(X)\le n$ for all $X\in S$. Note that there are $n^{2012}$ different ways to determine the values of $f(X)$ for all sets $X$ in $S$ of size $2011$. We will prove that the values of $f$ for the remaining sets are uniquely determined.

We apply induction on $k$ to prove that $$f(A\setminus \{a_1,a_2,\dots ,a_k\})=\min \{f(A\setminus \{a_1\}),f(A\setminus \{a_2\}),\dots ,f(A\setminus \{a_k\})\}(\ast )$$ for all $a_1,a_2,\dots ,a_k\in B$.

For $k=1$, it is trivial and the case for $k=2$ results from the condition on $f$.

$$f(A\setminus \{a_1,a_2,\dots ,a_{k+1}\})=\min \{f(A\setminus \{a_1,a_2,\dots ,a_k\}),f(A\setminus \{a_{k+1}\})\}.$$

Using the induction hypothesis and the observation above we can conclude that the claim is true for $k+1$ as well.

On the other hand, by (*) and the observation above it can be easily verified that the condition on $f$ is satisfied for all $A_1,A_2\in S$. Therefore, as $1\le n\le 19$ we obtain that there are $1^{2012}+2^{2012}+\cdots +19^{2012}$ such functions.