# 2013 AIME II Problems/Problem 11

## Problem 11

Let $A = \{1, 2, 3, 4, 5, 6, 7\}$, and let $N$ be the number of functions $f$ from set $A$ to set $A$ such that $f(f(x))$ is a constant function. Find the remainder when $N$ is divided by $1000$.

## Solution 1

Any such function can be constructed by distributing the elements of $A$ on three tiers.

The bottom tier contains the constant value, $c=f(f(x))$ for any $x$. (Obviously $f(c)=c$.)

The middle tier contains $k$ elements $x\ne c$ such that $f(x)=c$, where $1\le k\le 6$.

The top tier contains $6-k$ elements such that $f(x)$ equals an element on the middle tier.

There are $7$ choices for $c$. Then for a given $k$, there are $\tbinom6k$ ways to choose the elements on the middle tier, and then $k^{6-k}$ ways to draw arrows down from elements on the top tier to elements on the middle tier.

Thus $N=7\cdot\sum_{k=1}^6\tbinom6k\cdot k^{6-k}=7399$, giving the answer $\boxed{399}$.

## Solution 1 Clarified

Define the three layers as domain $x$, codomain $f(x)$, and codomain $f(f(x))$. Each one of them is contained in the set $A$. We know that $f(f(x))$ is a constant function, or in other words, can only take on one value. So, we can start off by choosing that value $c$ in $7$ ways. So now, we choose the values that can be $f(x)$ for all those values should satisfy $f(f(x))=c$. Let's $S$ be that set of values. First things first, we must have $5$ to be part of $S$, for the $S$ is part of the domain of $x$. Since the values in $i\in S$ all satisfy $f(i) = 5$, we have $5$ to be a value that $f(x)$ can be. Now, for the elements other than $5$:

If we have $k$ elements other than $5$ that can be part of $S$, we will have $\binom{6}{k}$ ways to choose those values. There will also be $k$ ways for each of the elements in $A$ other than $5$ and those in set $S$ (for when function $f$ is applied on those values, we already know it would be $5$). There are $6-k$ elements in $A$ other than $5$ and those in set $S$. So, there should be $6^{6-k}$ ways to match the domain $x$ to the values of $f(x)$. So, summing up all possible values of $k$ ( $[1,6]$), we have $$\sum_{k=1}^6 k^{6-k} = 6\cdot 1 + 15\cdot 16 + 20\cdot 27 + 15\cdot 16 + 6\cdot 5 + 1) = 1057$$

Multiplying that by the original $7$ for the choice of $c$, we have $7 \cdot 1057 = 7\boxed{399}.$

~sml1809

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