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

(→‎Solution ?)
(→‎Solution ?)
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*summing the cases
 
*summing the cases
−
we get that the probability for <math>f</math> and <math>g</math> to have disjoint ranges is equal to:
+
 
 +
we get that the probability for <math>f</math> and <math>g</math> to have 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{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>
 
so the final answer is <math>645</math>
 
so the final answer is <math>645</math>

Revision as of 16:52, 14 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 ?

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

so there $4*3^4$ 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)

or $g$ 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

here are 4 ways to choose the range of $f$ ${{4}\choose {1}}$ then each element in $A$ can be assigned to only one value so there are 4 functions with a range of 1 elements 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. so there are $4*(3^4+3*2^4+3)$ ways for that to occur

  • summing the cases

we get that the probability for $f$ and $g$ to have disjoint ranges is equal to:

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

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