Difference between revisions of "2010 AIME II Problems/Problem 8"
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− | == Problem | + | == Problem == |
− | Let <math>N</math> be the number of ordered | + | Let <math>N</math> be the number of [[ordered pair]]s of nonempty sets <math>\mathcal{A}</math> and <math>\mathcal{B}</math> that have the following properties: |
<UL> | <UL> | ||
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Find <math>N</math>. | Find <math>N</math>. | ||
− | == | + | == Solution== |
− | + | Let us [[partition]] the set <math>\{1,2,\cdots,12\}</math> into <math>n</math> numbers in <math>A</math> and <math>12-n</math> numbers in <math>B</math>, | |
− | Let us partition the set <math>\{1,2,\cdots,12\}</math> into <math>n</math> numbers in <math>A</math> and <math>12-n</math> numbers in <math>B</math>, | ||
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− | |||
+ | Since <math>n</math> must be in <math>B</math> and <math>12-n</math> must be in <math>A</math> (<math>n\ne6</math>, we cannot partition into two sets of 6 because <math>6</math> needs to end up somewhere, <math>n\ne 0</math> or <math>12</math> either). | ||
We have <math>\dbinom{10}{n-1}</math> ways of picking the numbers to be in <math>A</math>. | We have <math>\dbinom{10}{n-1}</math> ways of picking the numbers to be in <math>A</math>. | ||
− | So the answer is <math>\left(\sum_{n=1}^{11} \dbinom{10}{n-1}\right) - \dbinom{10}{5}=2^{10}-252= \boxed{772}</math> | + | So the answer is <math>\left(\sum_{n=1}^{11} \dbinom{10}{n-1}\right) - \dbinom{10}{5}=2^{10}-252= \boxed{772}</math>. |
== See also == | == See also == | ||
{{AIME box|year=2010|num-b=7|num-a=9|n=II}} | {{AIME box|year=2010|num-b=7|num-a=9|n=II}} | ||
+ | |||
+ | [[Category:Intermediate Combinatorics Problems]] |
Revision as of 11:46, 6 April 2010
Problem
Let be the number of ordered pairs of nonempty sets
and
that have the following properties:
-
,
-
,
- The number of elements of
is not an element of
,
- The number of elements of
is not an element of
.
Find .
Solution
Let us partition the set into
numbers in
and
numbers in
,
Since must be in
and
must be in
(
, we cannot partition into two sets of 6 because
needs to end up somewhere,
or
either).
We have ways of picking the numbers to be in
.
So the answer is .
See also
2010 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |