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Difference between revisions of "2010 AIME II Problems/Problem 8"

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== Problem 8 ==
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== Problem ==
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Let <math>N</math> be the number of ordered pairs of nonempty sets <math>\mathcal{A}</math> and <math>\mathcal{B}</math> that have the following properties:
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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:
  
 
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<UL>
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Find <math>N</math>.
 
Find <math>N</math>.
  
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== solution==
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== Solution==
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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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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)
 
  
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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>.  
  
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So the answer is <math>\left(\sum_{n=1}^{11} \dbinom{10}{n-1}\right) - \dbinom{10}{5}=2^{10}-252= \boxed{772}</math>
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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}}
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[[Category:Intermediate Combinatorics Problems]]

Revision as of 11:46, 6 April 2010

Problem

Let $N$ be the number of ordered pairs of nonempty sets $\mathcal{A}$ and $\mathcal{B}$ that have the following properties:

  • $\mathcal{A} \cup \mathcal{B} = \{1,2,3,4,5,6,7,8,9,10,11,12\}$,
  • $\mathcal{A} \cap \mathcal{B} = \emptyset$,
  • The number of elements of $\mathcal{A}$ is not an element of $\mathcal{A}$,
  • The number of elements of $\mathcal{B}$ is not an element of $\mathcal{B}$.

Find $N$.

Solution

Let us partition the set $\{1,2,\cdots,12\}$ into $n$ numbers in $A$ and $12-n$ numbers in $B$,

Since $n$ must be in $B$ and $12-n$ must be in $A$ ($n\ne6$, we cannot partition into two sets of 6 because $6$ needs to end up somewhere, $n\ne 0$ or $12$ either).

We have $\dbinom{10}{n-1}$ ways of picking the numbers to be in $A$.

So the answer is $\left(\sum_{n=1}^{11} \dbinom{10}{n-1}\right) - \dbinom{10}{5}=2^{10}-252= \boxed{772}$.

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
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