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Difference between revisions of "2014 AIME I Problems/Problem 12"

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m (→‎Solution)
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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 ==
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== Solution (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 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.
  
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*Case 1: <math>f</math>'s range contains 1 element
 
*Case 1: <math>f</math>'s range contains 1 element
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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>f</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>.
+
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>.
  
 
*Case 2: <math>f</math>'s range contains 2 elements
 
*Case 2: <math>f</math>'s range contains 2 elements

Revision as of 12:44, 15 March 2014

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 (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 $\boxed{453}$.

See also

2014 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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