2014 AIME I Problems/Problem 12

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

We note there are $4^8$ sets of two functions $f$ and $g$ from $A$ to $A$ since the input of the four values of each function has four options each. By the pigeonhole principle, the combined range of $f$ and $g$ has at most four elements this can be done in 3 cases:

Case 1: $f$ 's range contains 3 elements

There are 4 ways to choose the range of $f$ ${{4}\choose {3}}$ then each element in $f$ can be assigned to one of the 3 elements in the range of $f$ so there are $4*3^4$ functions with a range of 3 elements in $A$ .

Note that $g$ can only be the function assigning each element of $A$ to the element not already chosen.

Thus, there are $4*3^4$ ways for that to happen.

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

there are 6 ways to choose the range of $f$ ${{4}\choose {2}}$ then each element in $f$ can be assigned to one of the 2 elements in the range of $f$ so there are $4*2^4$ functions with a range of 3 elements in $A$ .

Now $g$ can have a range of two elements in which each of its 4 values can be assigned one of its elements ( $2^4$ ways)

Also, $g$ can have a range of one element there are 2 ways to choose its range ${{2}\choose {1}}$ , and like in case 1, $g$ can happen in exactly 1 way.

So there $(6*2^4)(2^4+2)$ ways for that to happen.

Case 3: $f$ 's range contains 1 element

There are 4 ways to choose the range of $f$ , given by ${{4}\choose {1}}$ . We then know that each element in $A$ can be assigned to only one value, so there are 4 functions with a range of 1 element in $A$ .

Now $g$ can have a range of 1, 2, or 3: If $g$ 's range has 3 elements each value in $A$ can be assigned to 3 other values so there $3^4$ ways for that to occur.

If $g$ 's range contains 2 elements one can choose the 2 elements in 3 different ways ${{3}\choose {2}}$ and after choosing each element has 2 options resulting with $2^4$ ways for that to occur

If $g$ 's range contains 1 element one can choose the 1 elements in 3 different ways ${{3}\choose {1}}$ and after choosing each element has 1 option resulting with 1 way for that to happen. Thus, there are $4*(3^4+3*2^4+3)$ ways for that to occur

Summing the cases

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

$\dfrac{4*3^4+(6*2^4)(2^4+2)+4*(3^4+3*2^4+3)}{4^8}=\dfrac{2^2*3^4+2^6*3^3+2^4*3*11}{2^{16}}= \dfrac{3^4+2^4*3^3+3*11*2^2}{2^{14}}=\dfrac{645}{2^{14}}$

Thus, our final answer is $645$ .

2014 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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2014 AIME I Problems/Problem 12

Problem 12

Solution

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