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Difference between revisions of "2018 AIME II Problems/Problem 10"

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Find the number of functions <math>f(x)</math> from <math>\{1, 2, 3, 4, 5\}</math> to <math>\{1, 2, 3, 4, 5\}</math> that satisfy <math>f(f(x)) = f(f(f(x)))</math> for all <math>x</math> in <math>\{1, 2, 3, 4, 5\}</math>.
 
Find the number of functions <math>f(x)</math> from <math>\{1, 2, 3, 4, 5\}</math> to <math>\{1, 2, 3, 4, 5\}</math> that satisfy <math>f(f(x)) = f(f(f(x)))</math> for all <math>x</math> in <math>\{1, 2, 3, 4, 5\}</math>.
  
 +
==Solution 1==
 +
 +
Just to visualize solution 1. If we list all possible <math>(x,f(x))</math>, from <math>{1,2,3,4,5}</math> to <math>{1,2,3,4,5}</math> in a specific order, we get <math>5*5 = 25</math> different <math>(x,f(x))</math> 's.
 +
Namely:
 +
 +
<math>(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)</math>
 +
 +
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 <math>(x,x)</math> where <math>x\in{1,2,3,4,5}</math> must exist.In this case I rather "go backwards". First fixing <math>5</math> pairs <math>(x,x)</math>, (the diagonal of our table) and map them to the other fitting pairs <math>(x,f(x))</math>. You can do this in <math>\frac{5!}{5!} = 1</math> way. Then fixing <math>4</math> pairs <math>(x,x)</math> (The diagonal minus <math>1</math>) and map them to the other fitting pairs <math>(x,f(x))</math>. You can do this in
 +
<math>4\cdot\frac{5!}{4!} = 20</math> ways. Then fixing <math>3</math> pairs <math>(x,x)</math> (The diagonal minus <math>2</math>) and map them to the other fitting pairs <math>(x,f(x))</math>. You can do this in <math>\frac{(5\cdot4\cdot3\cdot6\cdot3)}{3!2!} + \frac{(5\cdot4\cdot3\cdot6\cdot1)}{3!} = 150</math> ways.
 +
Fixing <math>2</math> pairs <math>(x,x)</math> (the diagonal minus <math>3</math>) and map them to the other fitting pairs <math>(x,f(x))</math>. You can do this in <math>\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</math> ways.
 +
Lastely, fixing <math>1</math> pair <math>(x,x)</math> (the diagonal minus <math>4</math>) and map them to the other fitting pairs <math>(x,f(x))</math>. You can do this in <math>\frac{5!}{4!} + 4\cdot\frac{5!}{3!} + 5! = 205</math>
 +
 +
So <math>1 + 20 + 150 + 380 + 205 = \framebox{756}</math>
 +
 +
==Solution 2==
 +
 +
We can do some caseworks about the special points of functions <math>f</math> for <math>x\in\{1,2,3,4,5\}</math>. Let <math>x</math>, <math>y</math> and <math>z</math> be three different elements in set <math>\{1,2,3,4,5\}</math>. There must be elements such like <math>k</math> in set <math>\{1,2,3,4,5\}</math> satisfies <math>f(k)=k</math>, and we call the points such like <math>(k,k)</math> on functions <math>f</math> 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 <math>3</math> because the highest iterations of function <math>f</math> is <math>3</math>. Now we can classify <math>3</math> cases of “Good points” of <math>f</math>.
 +
 +
<math>\textbf{Case 1:}</math> One "step" to "Good Points": Assume that <math>f(x)=x</math>, then we get <math>f(f(x))=f(x)=x</math>, and <math>f(f(f(x)))=f(f(x))=f(x)=x</math>, so <math>f(f(f(x)))=f(f(x))</math>.
 +
 +
<math>\textbf{Case 2:}</math> Two "steps" to "Good Points": Assume that <math>f(x)=y</math> and <math>f(y)=y</math>, then we get <math>f(f(x))=f(y)=y</math>, and <math>f(f(f(x)))=f(f(y))=f(y)=y</math>, so <math>f(f(f(x)))=f(f(x))</math>.
 +
 +
<math>\textbf{Case 3:}</math> Three "steps" to "Good Points": Assume that <math>f(x)=y</math>, <math>f(y)=z</math> and <math>f(z)=z</math>, then we get <math>f(f(x))=f(y)=z</math>, and <math>f(f(f(x)))=f(f(y))=f(z)=z</math>, so <math>f(f(f(x)))=f(f(x))</math>.
 +
 +
Divide set <math>\{1,2,3,4,5\}</math> into three parts which satisfy these three cases, respectively. Let the first part has <math>a</math> elements, the second part has <math>b</math> elements and the third part has <math>c</math> elements, it is easy to see that <math>a+b+c=5</math>. First, there are <math>\binom{5}{a}</math> ways to select <math>x</math> for Case 1. Second, we have <math>\binom{5-a}{b}</math> ways to select <math>x</math> 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 <math>a^b</math>. Finally, we map all the elements that satisfy Case 3 to Case 2, and the total number of ways of this operation is <math>b^c</math>. As a result, the number of such functions <math>f</math> can be represented in an algebraic expression contains <math>a</math>, <math>b</math> and <math>c</math>: <math>\boxed{\binom{5}{a}\cdot \binom{5-a}{b}\cdot a^b\cdot b^c}</math>
 +
 +
Now it's time to consider about the different values of <math>a</math>, <math>b</math> and <math>c</math> and the total number of functions <math>f</math> satisfy these values of <math>a</math>, <math>b</math> and <math>c</math>:
 +
 +
For <math>a=5</math>, <math>b=0</math> and <math>c=0</math>, the number of <math>f</math>s is <math>\binom{5}{5}=1</math>
 +
 +
For <math>a=4</math>, <math>b=1</math> and <math>c=0</math>, the number of <math>f</math>s is <math>\binom{5}{4}\cdot \binom{1}{1}\cdot 4^1\cdot 1^0=20</math>
 +
 +
For <math>a=3</math>, <math>b=1</math> and <math>c=1</math>, the number of <math>f</math>s is <math>\binom{5}{3}\cdot \binom{2}{1}\cdot 3^1\cdot 1^1=60</math>
 +
 +
For <math>a=3</math>, <math>b=2</math> and <math>c=0</math>, the number of <math>f</math>s is <math>\binom{5}{3}\cdot \binom{2}{2}\cdot 3^2\cdot 2^0=90</math>
 +
 +
For <math>a=2</math>, <math>b=1</math> and <math>c=2</math>, the number of <math>f</math>s is <math>\binom{5}{2}\cdot \binom{3}{1}\cdot 2^1\cdot 1^2=60</math>
 +
 +
For <math>a=2</math>, <math>b=2</math> and <math>c=1</math>, the number of <math>f</math>s is <math>\binom{5}{2}\cdot \binom{3}{2}\cdot 2^2\cdot 2^1=240</math>
 +
 +
For <math>a=2</math>, <math>b=3</math> and <math>c=0</math>, the number of <math>f</math>s is <math>\binom{5}{2}\cdot \binom{3}{3}\cdot 2^3\cdot 3^0=80</math>
 +
 +
For <math>a=1</math>, <math>b=1</math> and <math>c=3</math>, the number of <math>f</math>s is <math>\binom{5}{1}\cdot \binom{4}{1}\cdot 1^1\cdot 1^3=20</math>
 +
 +
For <math>a=1</math>, <math>b=2</math> and <math>c=2</math>, the number of <math>f</math>s is <math>\binom{5}{1}\cdot \binom{4}{2}\cdot 1^2\cdot 2^2=120</math>
 +
 +
For <math>a=1</math>, <math>b=3</math> and <math>c=1</math>, the number of <math>f</math>s is <math>\binom{5}{1}\cdot \binom{4}{3}\cdot 1^3\cdot 3^1=60</math>
 +
 +
For <math>a=1</math>, <math>b=4</math> and <math>c=0</math>, the number of <math>f</math>s is <math>\binom{5}{1}\cdot \binom{4}{4}\cdot 1^4\cdot 4^0=5</math>
 +
 +
Finally, we get the total number of function <math>f</math>, the number is <math>1+20+60+90+60+240+80+20+120+60+5=\boxed{756}</math>
 +
 +
~Solution by <math>BladeRunnerAUG</math> (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
 +
 +
==See Also==
 
{{AIME box|year=2018|n=II|num-b=9|num-a=11}}
 
{{AIME box|year=2018|n=II|num-b=9|num-a=11}}
−
{{MAA Notice}}
 
−
 
−
==Solution==
 
−
{{AIME box|year=2018|n=II|num-b=10|num-a=12}}
 
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 20:34, 23 August 2020

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 $5*5 = 25$ 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 $\frac{(5\cdot4\cdot3\cdot6\cdot3)}{3!2!} + \frac{(5\cdot4\cdot3\cdot6\cdot1)}{3!} = 150$ 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. Lastely, 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 $\frac{5!}{4!} + 4\cdot\frac{5!}{3!} + 5! = 205$

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

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 $f$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

See Also

2018 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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