Harvard-MIT Mathematics Tournament

Solution:
Answer: $2^{2011}-2012$

Let $n$ be the element of $A$ not in the range of $f$. Let $m$ be the element of $A$ that is hit twice.

We now sum the total number of functions over $n, m$. Clearly $f(1)=1$, and by induction, for $x \leq m$, $f(x)=x$. Also unless $n=2011$, $f(2011)=2011$ because $f$ can take no other number to $2011$. It follows from backwards induction that for $x>n$, $f(x)=x$. Therefore $n>m$, and there are only $n-m$ values of $f$ that are not fixed.

Now $f(m+1)=m$ or $f(m+1)=m+1$. For $m<k<n$, given the selection of $f(1), f(2), \ldots, f(k-1)$, $k-1$ of the $k+1$ possible values of $f(k+1)$ ($1,2,3, \ldots, k$, and counting $m$ twice) have been taken, so there are two distinct values that $f(k+1)$ can take (one of them is $k+1$, and the other is not, so they are distinct). For $f(n)$, when the other 2010 values of $f$ have been assigned, there is only one missing, so $f(n)$ is determined.

For each integer in $[m, n)$, there are two possible values of $f$, so there are $2^{n-m-1}$ different functions $f$ for a given $m, n$. So our answer is

$$\begin{aligned} \sum_{m=1}^{2010} \sum_{n=m+1}^{2011} 2^{n-m-1} & =\sum_{m=1}^{2010} 2^{-m-1} \sum_{n=m+1}^{2011} 2^{n} \\ & =\sum_{m=1}^{2010} 2^{-m-1}\left(2^{2012}-2^{m+1}\right) \\ & =\sum_{m=1}^{2010} 2^{2011-m}-1 \\ & =\left(\sum_{m=1}^{2010} 2^{m}\right)-2010 \\ & =2^{2011}-2012 \end{aligned}$$