# 2018 AIME II Problems/Problem 10

## Problem

Find the number of functions $f(x)$ from $\{1, 2, 3, 4, 5\}$ to $\{1, 2, 3, 4, 5\}$ that satisfy $f(f(x)) = f(f(f(x)))$ for all $x$ in $\{1, 2, 3, 4, 5\}$.

## Solution 1

Just to visualize solution 1. If we list all possible $(x,f(x))$, from ${1,2,3,4,5}$ to ${1,2,3,4,5}$ in a specific order, we get $5 \cdot 5 = 25$ different $(x,f(x))$ 's. Namely:

$$(1,1) (1,2) (1,3) (1,4) (1,5)$$ $$(2,1) (2,2) (2,3) (2,4) (2,5)$$ $$(3,1) (3,2) (3,3) (3,4) (3,5)$$ $$(4,1) (4,2) (4,3) (4,4) (4,5)$$ $$(5,1) (5,2) (5,3) (5,4) (5,5)$$

To list them in this specific order makes it a lot easier to solve this problem. We notice, In order to solve this problem at least one pair of $(x,x)$ where $x\in{1,2,3,4,5}$ must exist.In this case I rather "go backwards". First fixing $5$ pairs $(x,x)$, (the diagonal of our table) and map them to the other fitting pairs $(x,f(x))$. You can do this in $\frac{5!}{5!} = 1$ way. Then fixing $4$ pairs $(x,x)$ (The diagonal minus $1$) and map them to the other fitting pairs $(x,f(x))$. You can do this in $4\cdot\frac{5!}{4!} = 20$ ways. Then fixing $3$ pairs $(x,x)$ (The diagonal minus $2$) and map them to the other fitting pairs $(x,f(x))$. You can do this in $\tfrac{(5\cdot4\cdot3\cdot6\cdot3)}{3!2!} + \tfrac{(5\cdot4\cdot3\cdot6\cdot1)}{3!} = 150$ ways. Fixing $2$ pairs $(x,x)$ (the diagonal minus $3$) and map them to the other fitting pairs $(x,f(x))$. You can do this in $\frac{(5\cdot4\cdot6\cdot4\cdot2)}{2!3!} + \frac{(5\cdot4\cdot6\cdot4\cdot2)}{2!2!} + \frac{(5\cdot4\cdot6\cdot2\cdot1)}{2!2!} = 380$ ways. Lastly, fixing $1$ pair $(x,x)$ (the diagonal minus $4$) and map them to the other fitting pairs $(x,f(x))$. You can do this in $\tfrac{5!}{4!} + 4\cdot\tfrac{5!}{3!} + 5! = 205$ ways.

So $1 + 20 + 150 + 380 + 205 = \framebox{756}$

## Solution 2

We perform casework on the number of fixed points (the number of points where $f(x) = x$). To better visualize this, use the grid from Solution 1.

Case 1: 5 fixed points

- Obviously, there must be $1$ way to do so.

Case 2: 4 fixed points

- $\binom 54$ ways to choose the $4$ fixed points. For the sake of conversation, let them be $(1, 1), (2, 2), (3, 3), (4, 4)$.
- The last point must be $(5, 1), (5, 2), (5, 3),$ or $(5, 4)$.
- There are $\binom 54 \cdot 4 = 20$ solutions for this case.

Case 3: 3 fixed points

- $\binom 53$ ways to choose the $3$ fixed points. For the sake of conversation, let them be $(1, 1), (2, 2), (3, 3)$.
- Subcase 3.1: None of $4$ or $5$ map to each other
- The points must be $(4, 1), (4, 2), (4, 3)$ and $(5, 1), (5, 2), (5, 3)$, giving $3 \cdot 3 = 9$ solutions.
- Subcase 3.2: $4$ maps to $5$ or $5$ maps to $4$
- The points must be $(4, 5)$ and $(5, 1), (5, 2), (5, 3)$ or $(5, 4)$ and $(4, 1), (4, 2), (4, 3)$, giving $3+3 = 6$ solutions.
- There are $\binom 53 \cdot (9+6) = 150$ solutions for this case.

Case 4: 2 fixed points

- $\binom 52$ ways to choose the $2$ fixed points. For the sake of conversation, let them be $(1, 1), (2, 2)$.
- Subcase 4.1: None of $3$, $4$, or $5$ map to each other
- There are $2 \cdot 2 \cdot 2 = 8$ solutions for this case, by similar logic to Subcase 3.1.
- Subcase 4.2: exactly one of $3, 4, 5$ maps to another.
- Example: $(3, 4), (4, 1), (5, 2)$
- $\binom 32 \cdot 2! = 6$ ways to choose the 2 which map to each other, and $2\cdot 2$ ways to choose which ones of $1$ and $2$ they map to, giving $24$ solutions for this case.
- Subcase 4.3: two of $3, 4, 5$ map to the third
- Example: $(3, 5), (4, 5), (5, 2)$
- $\binom 31$ ways to choose which point is being mapped to, and $2$ ways to choose which one of $1$ and $2$ it maps to, giving $6$ solutions for this case.
- There are $\binom 52 \cdot (8+24+6) = 380$ solutions.

Case 5: 1 fixed point

- $\binom 51$ ways to choose the fixed point. For the sake of conversation, let it be $(1, 1)$.
- Subcase 5.1: None of $2, 3, 4, 5$ map to each other
- Obviously, there is $1^4 = 1$ solution to this; all of them map to $1$.
- Subcase 5.2: One maps to another, and the other two map to $1$.
- Example: $(2, 3), (3, 1), (4, 1), (5, 1)$
- There are $\binom 42 \cdot 2!$ ways to choose which two map to each other, and since each must map to $1$, this gives $12$.
- Subcase 5.3: One maps to another, and of the other two, one maps to the other as well.
- Example: $(2, 3), (3, 1), (5, 4), (4, 1)$
- There are $\binom 42 \cdot 2! \cdot 2! / 2!$ ways to choose which ones map to another. This also gives $12$.
- Subcase 5.4: 2 map to a third, and the fourth maps to $1$.
- Example: $(4, 2), (5, 2), (2, 1), (3, 1)$
- There are $\binom 42 \cdot \binom 21 = 12$ ways again.
- Subcase 5.5: 3 map to the fourth.
- Example: $(2, 4), (3, 4), (5, 4), (4, 1)$
- There are $\binom 41$ ways to choose which one is being mapped to, giving $4$ solutions.
- There are $\binom 51 \cdot (1+12+12+12+4) = 205$ solutions.

Therefore, the answer is $1+20+150+380+205 = \boxed{756}$

~First

## Solution 3

We can do some caseworks about the special points of functions $f$ for $x\in\{1,2,3,4,5\}$. Let $x$, $y$ and $z$ be three different elements in set $\{1,2,3,4,5\}$. There must be elements such like $k$ in set $\{1,2,3,4,5\}$ satisfies $f(k)=k$, and we call the points such like $(k,k)$ on functions $f$ are "Good Points" (Actually its academic name is "fixed-points"). The only thing we need to consider is the "steps" to get "Good Points". Notice that the "steps" must less than $3$ because the highest iterations of function $f$ is $3$. Now we can classify $3$ cases of “Good points” of $f$.

$\textbf{Case 1:}$ One "step" to "Good Points": Assume that $f(x)=x$, then we get $f(f(x))=f(x)=x$, and $f(f(f(x)))=f(f(x))=f(x)=x$, so $f(f(f(x)))=f(f(x))$.

$\textbf{Case 2:}$ Two "steps" to "Good Points": Assume that $f(x)=y$ and $f(y)=y$, then we get $f(f(x))=f(y)=y$, and $f(f(f(x)))=f(f(y))=f(y)=y$, so $f(f(f(x)))=f(f(x))$.

$\textbf{Case 3:}$ Three "steps" to "Good Points": Assume that $f(x)=y$, $f(y)=z$ and $f(z)=z$, then we get $f(f(x))=f(y)=z$, and $f(f(f(x)))=f(f(y))=f(z)=z$, so $f(f(f(x)))=f(f(x))$.

Divide set $\{1,2,3,4,5\}$ into three parts which satisfy these three cases, respectively. Let the first part has $a$ elements, the second part has $b$ elements and the third part has $c$ elements, it is easy to see that $a+b+c=5$. First, there are $\binom{5}{a}$ ways to select $x$ for Case 1. Second, we have $\binom{5-a}{b}$ ways to select $x$ for Case 2. After that we map all elements that satisfy Case 2 to Case 1, and the total number of ways of this operation is $a^b$. Finally, we map all the elements that satisfy Case 3 to Case 2, and the total number of ways of this operation is $b^c$. As a result, the number of such functions $f$ can be represented in an algebraic expression contains $a$, $b$ and $c$: $\boxed{\binom{5}{a}\cdot \binom{5-a}{b}\cdot a^b\cdot b^c}$

Now it's time to consider about the different values of $a$, $b$ and $c$ and the total number of functions $f$ satisfy these values of $a$, $b$ and $c$:

For $a=5$, $b=0$ and $c=0$, the number of $f$s is $\binom{5}{5}=1$

For $a=4$, $b=1$ and $c=0$, the number of $f$s is $\binom{5}{4}\cdot \binom{1}{1}\cdot 4^1\cdot 1^0=20$

For $a=3$, $b=1$ and $c=1$, the number of $f$s is $\binom{5}{3}\cdot \binom{2}{1}\cdot 3^1\cdot 1^1=60$

For $a=3$, $b=2$ and $c=0$, the number of $f$s is $\binom{5}{3}\cdot \binom{2}{2}\cdot 3^2\cdot 2^0=90$

For $a=2$, $b=1$ and $c=2$, the number of $f$s is $\binom{5}{2}\cdot \binom{3}{1}\cdot 2^1\cdot 1^2=60$

For $a=2$, $b=2$ and $c=1$, the number of $f$s is $\binom{5}{2}\cdot \binom{3}{2}\cdot 2^2\cdot 2^1=240$

For $a=2$, $b=3$ and $c=0$, the number of $f$s is $\binom{5}{2}\cdot \binom{3}{3}\cdot 2^3\cdot 3^0=80$

For $a=1$, $b=1$ and $c=3$, the number of $f$s is $\binom{5}{1}\cdot \binom{4}{1}\cdot 1^1\cdot 1^3=20$

For $a=1$, $b=2$ and $c=2$, the number of $f$s is $\binom{5}{1}\cdot \binom{4}{2}\cdot 1^2\cdot 2^2=120$

For $a=1$, $b=3$ and $c=1$, the number of $f$s is $\binom{5}{1}\cdot \binom{4}{3}\cdot 1^3\cdot 3^1=60$

For $a=1$, $b=4$ and $c=0$, the number of $f$s is $\binom{5}{1}\cdot \binom{4}{4}\cdot 1^4\cdot 4^0=5$

Finally, we get the total number of function $f$, the number is $1+20+60+90+60+240+80+20+120+60+5=\boxed{756}$

~Solution by $BladeRunnerAUG$ (Frank FYC)

## Note (fun fact)

This exact problem showed up earlier on the 2011 Stanford Math Tournament, Advanced Topics Test. This problem also showed up on the 2010 Mock AIME 2 here: https://artofproblemsolving.com/wiki/index.php/Mock_AIME_2_2010_Problems