# Difference between revisions of "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

Let the range be $R=\{f(x)|x \in A\}$

Let the constant function be $f(f(x))=c$, clearly, $c \in R$

For every $a \in R$, since $f(f(x))=c$, we must have $f(a)=c$.

Now we enumerate through possible sizes of $R$. $|R|$ cannot be more than 4 since all the numbers in $R$ must map to $c$ and any number other than $c$ in $R$ must have at least one number mapped to it.

1. $|R|=1$

We have $\binom{7}{1}=7$ choices for $R$, $\binom{1}{1}=1$ choices for $c$, and only 1 way to map the 6 numbers that are not in $R$ (remaining numbers) since the range $R$ is size 1. In total we have $7\cdot 1\cdot 1=7$ choices.

2. $|R|=2$

We have $\binom{7}{2}=21$ choices for $R$, $\binom{2}{1}=2$ choices for $c$.

Assigning the 5 numbers to the two elements in $R$ yields $2^5=32$ choices. However, one of these choices is to assign all 5 remaining numbers to $c$, resulting in $R$ being only of size 1. Exclude that and we have $32-1=31$ choices.

In total we have $21*2*31=1302$ choices.

3. $|R|=3$

We have $\binom{7}{3}=35$ choices for $R$, $\binom{3}{1}=3$ choices for $c$.

Notice that assigning the remaining numbers is equivalent to distributing 4 numbers into three groups where two of the three groups must receive at least 1 number (since we already have all numbers in $R$ mapped to $c$, the restriction is not necessary for the $c$ group). Ignoring the restrictions we have $3^4$ ways. Now minus the two cases where one of the two restricted group is left empty. There are 2 ways to choose the left-out group, and $2^4$ ways to distribute numbers for each choice. Finally, by inclusion-exclusion principle, we need to add back the case in which both restricted groups are left empty, which has only $1^4=1$ occurrence. In all, we have $3^4-2*2^4+1^4=50$ ways to assign the remaining numbers.

In total, we have $35*3*50=5250$ choices for this case.

4. $|R|=4$

We have $\binom{7}{4}=35$ choices for $R$, $\binom{4}{1}=4$ choices for $c$.

Now we have 3 "groups" to fill and only 3 remaining numbers. Thus each group must receive exactly 1 number. However, we still have $3!=6$ ways to permute the mappings.

Thus, we have $35*4*6=840$ choices in total.

To summarize, we have $7+1302+5250+840=7399$ different mappings possible. So the answer is $\boxed{399}$.

## Solution 2

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 $\boxed{399}$.