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>. | ||
− | == Solution | + | == Solution == |
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 | 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 | ||
By the pigeonhole principle the combined range of <math>f</math> and <math>g</math> has at most four elements | By the pigeonhole principle the combined range of <math>f</math> and <math>g</math> has at most four elements | ||
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*Case 1: <math>f</math>'s range contains 3 elements | *Case 1: <math>f</math>'s range contains 3 elements | ||
− | there are 4 ways to choose the range of <math>f</math> <math>{{4}\choose {3}}</math> then | + | there are 4 ways to choose the range of <math>f</math> <math>{{4}\choose {3}}</math> then three elements of A can be assigned in <math>0.5*3*4*(3!)</math> ways: choose which element will be assigned to more than once (3) how three elements will be assigned to to three elements of A(6) which of the four elements will be assigned to the extra value, divided by 2 becuase of over-counting |
− | + | ||
note that <math>g</math> can only be the function assigning each element of <math>A</math> to the element not already chosen | note that <math>g</math> can only be the function assigning each element of <math>A</math> to the element not already chosen | ||
− | so there <math>4* | + | so there <math>4*36</math> for that to happen |
*Case 2: <math>f</math>'s range contains 2 elements | *Case 2: <math>f</math>'s range contains 2 elements | ||
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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> | 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> | ||
− | 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) | + | 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>2^4-2</math> ways) |
or <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 | or <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 | ||
− | so there <math>(6*2^4)(2^4 | + | so there <math>(6*(2^4-2))(2^4)</math> ways for that to happen |
*case 3: <math>f</math>'s range contains 1 element | *case 3: <math>f</math>'s range contains 1 element | ||
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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. | 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. | ||
− | so there are <math>4*(3^4 | + | so there are <math>4*(3^4)</math> ways for that to occur |
*summing the cases | *summing the cases | ||
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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 | + | <math>\dfrac{4*3^4+(6*2^4)(2^4-2)+4*(36)}{4^8} =\dfrac{2^2*3^4+2^6*3*7+2^4*3^2}{2^{16}}= \dfrac{3^4+2^4*3*7+3^2*2^2}{2^{14}}=\dfrac{453}{2^{14}}</math> |
− | so the final answer is <math> | + | so the final answer is <math>453</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}} |
Revision as of 21:11, 14 March 2014
Problem 12
Let , and
and
be randomly chosen (not necessarily distinct) functions from
to
. The probability that the range of
and the range of
are disjoint is
, where
and
are relatively prime positive integers. Find
.
Solution
We note there are sets of two functions
and
from
to
since the input of the four values of each function has four options each
By the pigeonhole principle the combined range of
and
has at most four elements
this can be done in 3 cases:
- Case 1:
's range contains 3 elements
there are 4 ways to choose the range of
then three elements of A can be assigned in
ways: choose which element will be assigned to more than once (3) how three elements will be assigned to to three elements of A(6) which of the four elements will be assigned to the extra value, divided by 2 becuase of over-counting
note that can only be the function assigning each element of
to the element not already chosen
so there for that to happen
- Case 2:
's range contains 2 elements
there are 6 ways to choose the range of
then each element in
can be assigned to one of the 2 elements in the range of
so there are
functions with a range of 3 elements in
now can have a range of two elements in which each of its 4 values can be assigned one of its elements (
ways)
or can have a range of one element there are 2 ways to choose its range
and like in case 1
can happen in exactly 1 way
so there ways for that to happen
- case 3:
's range contains 1 element
here are 4 ways to choose the range of
then each element in
can be assigned to only one value so there are 4 functions with a range of 1 elements in
now can have a range of 1,2 or 3:
if
's range has 3 elements each value in
can be assigned to 3 other values so there
ways for that to occur.
if 's range contains 2 elements one can choose the 2 elements in 3 different ways
and after choosing each element has 2 options resulting with
ways for that to occur
if 's range contains 1 element one can choose the 1 elements in 3 different ways
and after choosing each element has 1 option resulting with 1 way for that to happen.
so there are
ways for that to occur
- summing the cases
we get that the probability for and
to have disjoint ranges is equal to:
so the final answer is
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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