Difference between revisions of "2014 AIME I Problems/Problem 12"
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<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> | ||
+ | |||
+ | == See also == | ||
+ | {{AIME box|year=2014|n=I|num-b=11|num-a=13}} | ||
+ | {{MAA Notice}} |
Revision as of 19:44, 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 each element in
can be assigned to one of the 3 elements in the range of
so there are
functions with a range of 3 elements in
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.