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Difference between revisions of "2014 AIME I Problems/Problem 12"
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Let <math>A=\{1,2,3,4\}</math>, and <math>f</math> and <math>g</math> be randomly chosen (not necessarily distinct) functions from <math>A</math> to <math>A</math>. The probability that the range of <math>f</math> and the range of <math>g</math> are disjoint is <math>\tfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m</math>.
 
Let <math>A=\{1,2,3,4\}</math>, and <math>f</math> and <math>g</math> be randomly chosen (not necessarily distinct) functions from <math>A</math> to <math>A</math>. The probability that the range of <math>f</math> and the range of <math>g</math> are disjoint is <math>\tfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m</math>.
  
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== Solution ==
+
==Solution 1 (casework)==
−
We note there are <math>4^8</math> sets of two functions <math>f</math> and <math>g</math> from <math>A</math> to <math>A</math> since the input of the four values of each function has four options each.
+
The natural way to go is casework. And the natural process is to sort <math>f</math> and <math>g</math> based on range size! Via Pigeonhole Principle, we see that the only real possibilities are: <math>f 1 g 1; f 1 g 2; f 1 g 3; f 2 g 2</math>. Note that the <math>1, 2</math> and <math>1, 3</math> cases are symmetrical and we need just a <math>*2</math>. Note also that the total number of cases is <math>4^4*4^4=4^8</math>.
−
By the pigeonhole principle, the combined range  of <math>f</math> and <math>g</math> has at most four elements
 
−
this can be done in 3 cases:
 
  
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*Case 1: <math>f</math>'s range contains 3 elements
+
<math>f 1 g 1</math>: clearly, we choose one number as the range for <math>f</math>, one for <math>g</math>, yielding <math>12</math> possibilities.
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There are 4 ways to choose the range of <math>f</math> <math>{{4}\choose {3}}</math> then each element in <math>f</math> can be assigned to one of the 3 elements in the range of <math>f</math> so there are <math>4*3^4</math> functions with a range of 3 elements in <math>A</math>.
 
  
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Note that <math>g</math> can only be the function assigning each element of <math>A</math> to the element not already chosen.
+
<math>f 1 g 2</math> with symmetry (WLOG <math>f</math> has 1 element): start by selecting numbers for the ranges. This yields <math>4</math> for the one number in <math>f</math>, and <math>3</math> options for the two numbers for <math>g</math>. Afterwards, note that the function with 2 numbers in the range can have <math>4+6+4=14</math> arrangements of these two numbers (1 of one, 3 of the other *2 and 2 of each). Therefore, we have <math>2*12*14</math> possibilities, the 2 from symmetry.
  
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Thus, there are <math>4*3^4</math> ways for that to happen.
+
<math>f 2 g 2</math>: no symmetry, still easy! Just note that we have <math>6</math> choices of which numbers go to <math>f</math> and <math>g</math>, and within each, <math>14*14=196</math> choices for the orientation of each of the two numbers. That's <math>6*196</math> possibilities.
  
−
*Case 2: <math>f</math>'s range contains 2 elements
+
<math>f 1 g 3</math>: again, symmetrical (WLOG <math>f</math> has one element): <math>4</math> ways to select the single element for <math>f</math>, and then find the number of ways to distribute the <math>3</math> distinct numbers in the range for <math>g</math>. The only arrangement for the frequency of each number is <math>{1, 1, 2}</math> in some order. Therefore, we have <math>3</math> ways to choose which number is the one represented twice, and then note that there are <math>12</math> ways to arrange these! The number of possibilities in this situation is <math>2 * 4 * 3 * 12</math>.
 +
 
 +
Total, divided by <math>4^8</math>, gets <math>\frac{3 * (1 + 2 * 7^2 + 2^2 * 7 + 2^3 * 3)}{4^7}</math>, with numerator <math>\boxed{453}</math>.
 +
 
 +
==Solution 2 (casework)==
 +
We note there are <math>4^4 = 256</math> possibilities for each of <math>f</math> and <math>g</math> from <math>A</math> to <math>A</math> since the input of the four values of each function has four options each for an output value.
 +
 
 +
We proceed with casework to determine the number of possible <math>f</math> with range 1, 2, etc.
 +
*Range 1:
 +
 
 +
There are 4 possibilities: all elements output to 1, 2, 3, or 4.
 +
 
 +
*Range 2:
 +
 
 +
We have <math>{{4}\choose {2}} = 6</math> ways to choose the two output elements for <math>f</math>. At this point we have two possibilities: either <math>f</math> has 3 of 1 element and 1 of the other, or 2 of each element. In the first case, there are 2 ways to pick the element which there are 3 copies of, and <math>{{4}\choose {1}} = 4</math> ways to rearrange the 4 elements, for a total of <math>6 * 4 * 2 = 48</math> ways for this case. For the second case, there are <math>{{4}\choose {2}} = 6</math> ways to rearrange the 4 elements, for a total of <math>6 * 6 = 36</math> ways for this case. Adding these two, we get a total of <math>36 + 48 = 84</math> total possibilities.
 +
 
 +
*Range 3:
 +
 
 +
We have <math>{{4}\choose {3}} = 4</math> ways to choose the three output elements for <math>f</math>. We know we must have 2 of 1 element and 1 of each of the others, so there are 3 ways to pick this element. Finally, there are <math>{{4}\choose{1}}*{{3}\choose{1}} = 12</math> ways to rearrange these elements (since we can pick the locations of the 2 single elements in this many ways), and our total is <math>4 * 3 * 12 = 144</math> ways.
  
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there are 6 ways to choose the range of <math>f</math> <math>{{4}\choose {2}}</math> then each element in <math>f</math> can be assigned to one of the 2 elements in the range of <math>f</math> so there are <math>4*2^4</math> functions with a range of 3 elements in <math>A</math>.
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*Range 4:
  
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Now <math>g</math> can have a range of two elements in which each of its 4 values can be assigned one of its elements (<math>2^4</math> ways)
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Since we know the elements present, we have <math>4!</math> ways to arrange them, or 24 ways.
  
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Also, <math>g</math> can have a range of one element there are 2 ways to choose its range <math>{{2}\choose {1}}</math>, and like in case 1, <math>g</math> can happen in exactly 1 way.
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(To check, <math>4 + 84 + 144 + 24 = 256</math>, which is the total number of possibilities).
  
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So there <math>(6*2^4)(2^4+2)</math> ways for that to happen.
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We now break <math>f</math> down by cases, and count the number of <math>g</math> whose ranges are disjoint from <math>f</math>'s.
  
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*Case 3: <math>f</math>'s range contains 1 element
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*Case 1: <math>f</math>'s range contains 1 element
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There are 4 ways to choose the range of <math>f</math>, given by <math>{{4}\choose {1}}</math>. We then know that each element in <math>A</math> can be assigned to only one value, so there are 4 functions with a range of 1 element in <math>A</math>.
+
We know that there are 3 possibilities for <math>g</math> with 1 element. Since half the possibilities for <math>g</math> with two elements will contain the element in <math>f</math>, there are <math>84/2 = 42</math> possibilities for <math>g</math> with 2 elements. Since <math>3/4</math> the possibilities for <math>g</math> with 3 elements will contain the element in <math>f</math>, there are <math>144/4 = 36</math> possibilities for <math>g</math> with 3 elements. Clearly, no 4-element range for <math>g</math> is possible, so the total number of ways for this case to happen is <math>4(3 + 42 + 36) = 324</math>.
  
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Now <math>g</math> can have a range of 1, 2, or 3:
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*Case 2: <math>f</math>'s range contains 2 elements
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If <math>g</math>'s range has 3 elements each value in <math>A</math> can be assigned to 3 other values so there <math>3^4</math> ways for that to occur.
+
We know that there are 2 possibilities for <math>g</math> with 1 element. If <math>g</math> has 2 elements in its range, they are uniquely determined, so the total number of sets with a range of 2 elements that work for <math>g</math> is <math>84/6 = 14</math>. No 3-element or 4-element ranges for <math>g</math> are possible. Thus, the total number of ways for this to happen is <math>84(2 + 14) = 1344</math>.
  
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If <math>g</math>'s range contains 2 elements one can choose the 2 elements in 3 different ways  <math>{{3}\choose {2}}</math> and after choosing each element has 2 options resulting with <math>2^4</math> ways for that to occur
 
  
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If <math>g</math>'s range contains 1 element one can choose the 1 elements in 3 different ways  <math>{{3}\choose {1}}</math> and after choosing each element has 1 option resulting with 1 way for that to happen.
+
*Case 3:  <math>f</math>'s range contains 3 elements
−
Thus, there are <math>4*(3^4+3*2^4+3)</math> ways for that to occur
+
In this case, there is only 1 possibility for <math>g</math> - all the output values are the element that does not appear in <math>f</math>'s range. Thus, the total number of ways for this to happen is <math>144</math>.
  
 
*Summing the cases
 
*Summing the cases
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We find that the probability of <math>f</math> and <math>g</math> having disjoint ranges is equal to:
 
We find that the probability of <math>f</math> and <math>g</math> having disjoint ranges is equal to:
  
−
<math>\dfrac{4*3^4+(6*2^4)(2^4+2)+4*(3^4+3*2^4+3)}{4^8}=\dfrac{2^2*3^4+2^6*3^3+2^4*3*11}{2^{16}}= \dfrac{3^4+2^4*3^3+3*11*2^2}{2^{14}}=\dfrac{645}{2^{14}}</math>
+
<math>\dfrac{324 + 1344 + 144}{256^2}=\dfrac{1812}{2^{16}}= \dfrac{453}{2^{14}}</math>
 +
 
 +
Thus, our final answer is <math>\boxed{453}</math>.
 +
 
 +
=== Solution 3 (simplification of Solution 2) ===
 +
As before, there are 4 functions with a range of size 1, 84 with a range of size 2, and 144 with a range of size 3. If the range of <math>f</math> has size <math>k</math>, the codomain of <math>g</math> is restricted to a set of size <math>4 - k</math>. Any function from <math>A</math> into this codomain will do, so there are <math>(4 - k)^4</math> possibilities for <math>g</math> given a function <math>f</math>. The probability of <math>f</math> and <math>g</math> having disjoint ranges is then
 +
<cmath>\frac{4\cdot 3^4 + 84\cdot 2^4 + 144\cdot 1^4}{(4^4)^2} = \frac{\boxed{453}}{2^{14}}.</cmath>
 +
 
 +
 
 +
==Video Solution==
 +
https://youtu.be/VFj6JesV93M?si=cDAOday30X0O__4T
 +
 
 +
~MathProblemSolvingSkills.com
  
−
Thus, our final answer is <math>645</math>.
 
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2014|n=I|num-b=11|num-a=13}}
 
{{AIME box|year=2014|n=I|num-b=11|num-a=13}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 20:31, 13 October 2023

Problem 12

Let $A=\{1,2,3,4\}$, and $f$ and $g$ be randomly chosen (not necessarily distinct) functions from $A$ to $A$. The probability that the range of $f$ and the range of $g$ are disjoint is $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m$.

Solution 1 (casework)

The natural way to go is casework. And the natural process is to sort $f$ and $g$ based on range size! Via Pigeonhole Principle, we see that the only real possibilities are: $f 1 g 1; f 1 g 2; f 1 g 3; f 2 g 2$. Note that the $1, 2$ and $1, 3$ cases are symmetrical and we need just a $*2$. Note also that the total number of cases is $4^4*4^4=4^8$.

$f 1 g 1$: clearly, we choose one number as the range for $f$, one for $g$, yielding $12$ possibilities.

$f 1 g 2$ with symmetry (WLOG $f$ has 1 element): start by selecting numbers for the ranges. This yields $4$ for the one number in $f$, and $3$ options for the two numbers for $g$. Afterwards, note that the function with 2 numbers in the range can have $4+6+4=14$ arrangements of these two numbers (1 of one, 3 of the other *2 and 2 of each). Therefore, we have $2*12*14$ possibilities, the 2 from symmetry.

$f 2 g 2$: no symmetry, still easy! Just note that we have $6$ choices of which numbers go to $f$ and $g$, and within each, $14*14=196$ choices for the orientation of each of the two numbers. That's $6*196$ possibilities.

$f 1 g 3$: again, symmetrical (WLOG $f$ has one element): $4$ ways to select the single element for $f$, and then find the number of ways to distribute the $3$ distinct numbers in the range for $g$. The only arrangement for the frequency of each number is ${1, 1, 2}$ in some order. Therefore, we have $3$ ways to choose which number is the one represented twice, and then note that there are $12$ ways to arrange these! The number of possibilities in this situation is $2 * 4 * 3 * 12$.

Total, divided by $4^8$, gets $\frac{3 * (1 + 2 * 7^2 + 2^2 * 7 + 2^3 * 3)}{4^7}$, with numerator $\boxed{453}$.

Solution 2 (casework)

We note there are $4^4 = 256$ possibilities for each of $f$ and $g$ from $A$ to $A$ since the input of the four values of each function has four options each for an output value.

We proceed with casework to determine the number of possible $f$ with range 1, 2, etc.

  • Range 1:

There are 4 possibilities: all elements output to 1, 2, 3, or 4.

  • Range 2:

We have ${{4}\choose {2}} = 6$ ways to choose the two output elements for $f$. At this point we have two possibilities: either $f$ has 3 of 1 element and 1 of the other, or 2 of each element. In the first case, there are 2 ways to pick the element which there are 3 copies of, and ${{4}\choose {1}} = 4$ ways to rearrange the 4 elements, for a total of $6 * 4 * 2 = 48$ ways for this case. For the second case, there are ${{4}\choose {2}} = 6$ ways to rearrange the 4 elements, for a total of $6 * 6 = 36$ ways for this case. Adding these two, we get a total of $36 + 48 = 84$ total possibilities.

  • Range 3:

We have ${{4}\choose {3}} = 4$ ways to choose the three output elements for $f$. We know we must have 2 of 1 element and 1 of each of the others, so there are 3 ways to pick this element. Finally, there are ${{4}\choose{1}}*{{3}\choose{1}} = 12$ ways to rearrange these elements (since we can pick the locations of the 2 single elements in this many ways), and our total is $4 * 3 * 12 = 144$ ways.

  • Range 4:

Since we know the elements present, we have $4!$ ways to arrange them, or 24 ways.

(To check, $4 + 84 + 144 + 24 = 256$, which is the total number of possibilities).

We now break $f$ down by cases, and count the number of $g$ whose ranges are disjoint from $f$'s.

  • Case 1: $f$'s range contains 1 element

We know that there are 3 possibilities for $g$ with 1 element. Since half the possibilities for $g$ with two elements will contain the element in $f$, there are $84/2 = 42$ possibilities for $g$ with 2 elements. Since $3/4$ the possibilities for $g$ with 3 elements will contain the element in $f$, there are $144/4 = 36$ possibilities for $g$ with 3 elements. Clearly, no 4-element range for $g$ is possible, so the total number of ways for this case to happen is $4(3 + 42 + 36) = 324$.

  • Case 2: $f$'s range contains 2 elements

We know that there are 2 possibilities for $g$ with 1 element. If $g$ has 2 elements in its range, they are uniquely determined, so the total number of sets with a range of 2 elements that work for $g$ is $84/6 = 14$. No 3-element or 4-element ranges for $g$ are possible. Thus, the total number of ways for this to happen is $84(2 + 14) = 1344$.


  • Case 3: $f$'s range contains 3 elements

In this case, there is only 1 possibility for $g$ - all the output values are the element that does not appear in $f$'s range. Thus, the total number of ways for this to happen is $144$.

  • Summing the cases

We find that the probability of $f$ and $g$ having disjoint ranges is equal to:

$\dfrac{324 + 1344 + 144}{256^2}=\dfrac{1812}{2^{16}}= \dfrac{453}{2^{14}}$

Thus, our final answer is $\boxed{453}$.

Solution 3 (simplification of Solution 2)

As before, there are 4 functions with a range of size 1, 84 with a range of size 2, and 144 with a range of size 3. If the range of $f$ has size $k$, the codomain of $g$ is restricted to a set of size $4 - k$. Any function from $A$ into this codomain will do, so there are $(4 - k)^4$ possibilities for $g$ given a function $f$. The probability of $f$ and $g$ having disjoint ranges is then \[\frac{4\cdot 3^4 + 84\cdot 2^4 + 144\cdot 1^4}{(4^4)^2} = \frac{\boxed{453}}{2^{14}}.\]


Video Solution

https://youtu.be/VFj6JesV93M?si=cDAOday30X0O__4T

~MathProblemSolvingSkills.com


See also

2014 AIME I (Problems • Answer Key • Resources)
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Problem 13
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