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

We do casework on the number of fixed points of $f$ , that is, the number of $x\in\{1,2,3,4,5\}$ such that $f(x)=x$ . We know that at least one such $x$ exists, namely $x=f(f(1))$ .

$\textbf{Case 1: one fixed point.}$
There are five ways to choose the fixed point. WLOG let the fixed point be $1$ . Then at least one of $2,3,4,5$ must map onto $1$ , the only fixed point.
Suppose exactly one of these values maps to $1$ ; there are four ways to choose such a value. WLOG let it be $2$ . Then all of $3,4,5$ must map to $2$ in order to be mapped to $1$ in the next iteration. There are $4$ solutions in this case.
Suppose exactly two of these values map to $1$ ; there are $\binom 4 2=6$ ways to choose such values. WLOG let them be $2$ and $3$ . Then $4$ and $5$ must map to one of $2$ and $3$ , where there are $2^2=4$ ways to do so. Therefore there are $6\cdot 4=24$ solutions in this case.
Suppose exactly three of these values map to $1$ ; there are $\binom 4 3=4$ ways to choose such values. WLOG let them be $2$ , $3$ , and $4$ . Then $5$ must map to one of $2$ , $3$ , and $4$ , where there are $3$ solutions. Therefore there are $4\cdot 3=12$ solutions in this case.
Suppose exactly four of these values map to $1$ . Then everything maps to $1$ and there is $1$ solution in this case.
Therefore there are $5\cdot(4+24+12+1)=205$ solutions in Case 1.

$\textbf{Case 2: two fixed points.}$
There are $\binom 5 2=10$ ways to choose the fixed points. WLOG let them be $1$ and $2$ . Then at least one of $3,4,5$ must map onto $1$ or $2$ .
Suppose exactly one of these values maps to $1$ or $2$ ; there are three ways to choose this value, and two ways to choose the value it maps to. WLOG let it be $3$ . Then both $4$ and $5$ must map to $3$ , for a total of $3\cdot 2=6$ solutions in this case.
Suppose exactly two of these values map to $1$ or $2$ ; there are $\binom 3 2=3$ ways to choose these values, and $2^2=4$ ways to choose the values they map to. WLOG let them be $3$ and $4$ . Then $5$ must map to one of $3$ and $4$ , for two possible choices. Therefore there are $3\cdot 2^2\cdot 2=24$ solutions in this case.
Suppose exactly three of these values map to $1$ or $2$ ; then everything maps to $1$ or $2$ and there are $2^3=8$ solutions in this case.
Therefore there are $10\cdot(6+24+8)=380$ solutions in Case 2.

$\textbf{Case 3: three fixed points.}$
There are $\binom 5 3=10$ ways to choose the fixed points. WLOG let them be $1$ , $2$ , and $3$ . Then at least one of $4$ and $5$ must map onto $1$ , $2$ , or $3$ .
Suppose exactly one of these values map to $1$ , $2$ , or $3$ ; there are two ways to choose this value, and three ways to choose the value is maps to. WLOG let it be $4$ . Then $5$ must map to $4$ , for a total of $2\cdot 3=6$ solutions in this case.
Suppose exactly two of these values map to $1$ , $2$ , or $3$ ; then everything maps to $1$ , $2$ , or $3$ , and there are $3^2=9$ solutions in this case.
Therefore there are $10\cdot(6+9)=150$ solutions in Case 3.

$\textbf{Case 4: four fixed points.}$
There are $\binom 5 4=5$ ways to choose the fixed points. WLOG let them to $1$ , $2$ , $3$ , and $4$ . Then $5$ must map to one of these values, for a total of $5\cdot 4=20$ solutions in Case 4.

$\textbf{Case 5: five fixed points.}$
Since everything is a fixed point, there is only one solution in Case 5.

Therefore there are a total of $205+380+150+20+1=\boxed{756}$ functions that satisfy the problem condition.

~Solution by ghghghghghghghgh

Solution 2

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 $f$ is $\binom{5}{5}=1$

For $a=4$ , $b=1$ and $c=0$ , the number of $f$ 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 $f$ 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 $f$ 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 $f$ 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 $f$ 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 $f$ 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 $f$ 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 $f$ 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 $f$ 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 $f$ 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 FYC

Note (fun fact)

This exact problem showed up earlier on the 2011 Stanford Math Tournament, Advanced Topics Test.

2018 AIME II (Problems • Answer Key • Resources)
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2018 AIME II Problems/Problem 10

Contents

Problem

Solution 1

Solution 2

Note (fun fact)