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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:
  
 
<UL>
 
<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> (if <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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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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Note: We have <math>\dbinom{10}{n-1}</math>  ways of picking the numbers to be in <math>A</math> because there are <math>n</math> numbers in <math>A</math> and since <math>12-n</math> is already a term in the set we simply have to choose another <math>n-1</math> numbers from the <math>10</math> numbers that are available.
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== Solution 2==
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Regardless of the size <math>n</math> of <math>A</math> (ignoring the case when <math>n = 6</math>), <math>n</math> must not be in <math>A</math> and <math>12 - n</math> must be in <math>A</math>.
 +
 
 +
There are <math>10</math> remaining elements whose placements have yet to be determined. Note that the actual value of <math>n</math> does not matter; there is always <math>1</math> necessary element, <math>1</math> forbidden element, and <math>10</math> other elements that need to be distributed. There are <math>2</math> places to put each of these elements, for <math>2^{10}</math> possibilities.
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 +
However, there is the edge case of <math>n = 6; 6</math> is forced not the be in either set, so we must subtract the <math>\dbinom{10}{5}</math> cases where <math>A</math> and <math>B</math> have size <math>6</math>.
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 +
Thus, our answer is <math>2^{10} - \dbinom{10}{5} = 1024 - 252 = \boxed{772}</math>
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 +
 
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== Solution 3 (PIE and Complementary Counting) ==
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The total number of possible subsets is <math>\sum_{i=1}^{11}\dbinom{12}{i}</math>, which is <math>2^{12}-2</math>. Note that picking a subset from the set leaves the rest of the set to be in the other subset. We exclude <math>i=0</math> and <math>i=12</math> since they leave a set empty. We proceed with complementary counting and casework:
 +
 
 +
We apply the Principle of Inclusion and Exclusion for casework on the complementary cases. We find the ways where <math>|A|</math> is in <math>A</math>, which violates the first condition. Then we find the ways where the elements <math>|B|</math> and <math>12-|B|</math> are in set <math>B</math>, which violates only the second condition, and not the first. This ensures we do not overcount.
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Case 1: <math>|A|</math> is an element in <math>A</math>
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There are <math>\sum_{i=1}^{11}\dbinom{11}{i-1}</math> = <math>2^{11}-1</math> ways in this case.
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Case 2: <math>|B|</math> and <math>12-|B|</math> are in <math>B</math>
  
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We introduce a subcase where <math>|B|</math> is not 6, since the other element would also be 6.
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There are <math>B-2</math> elements to choose from 10 elements. Therefore, <math>|B|</math> can be from 2 to 11.
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There are <math>\sum_{i=2}^{11}\dbinom{10}{i-2}-\dbinom{10}{4} = 2^{10}-211</math> ways.
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The other subcase is when <math>|B|</math> is equal to 6. There are <math>\dbinom{11}{5}=462</math> ways.
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Adding the subcases gives us <math>2^{10}+251</math>.
  
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We have <math>\dbinom{10}{n-1}</math> ways of picking the numbers to be in <math>A</math>.  
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Adding this with case one gives us <math>2^{11}+2^{10}+250</math>. Subtracting this from <math>2^{12}-2</math> gives <math>1024-252=\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>
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~[https://artofproblemsolving.com/wiki/index.php/User:Magnetoninja Magnetoninja]
  
 
== 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]]
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{{MAA Notice}}

Latest revision as of 19:57, 20 July 2023

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

Note: We have $\dbinom{10}{n-1}$ ways of picking the numbers to be in $A$ because there are $n$ numbers in $A$ and since $12-n$ is already a term in the set we simply have to choose another $n-1$ numbers from the $10$ numbers that are available.

Solution 2

Regardless of the size $n$ of $A$ (ignoring the case when $n = 6$), $n$ must not be in $A$ and $12 - n$ must be in $A$.

There are $10$ remaining elements whose placements have yet to be determined. Note that the actual value of $n$ does not matter; there is always $1$ necessary element, $1$ forbidden element, and $10$ other elements that need to be distributed. There are $2$ places to put each of these elements, for $2^{10}$ possibilities.

However, there is the edge case of $n = 6; 6$ is forced not the be in either set, so we must subtract the $\dbinom{10}{5}$ cases where $A$ and $B$ have size $6$.

Thus, our answer is $2^{10} - \dbinom{10}{5} = 1024 - 252 = \boxed{772}$


Solution 3 (PIE and Complementary Counting)

The total number of possible subsets is $\sum_{i=1}^{11}\dbinom{12}{i}$, which is $2^{12}-2$. Note that picking a subset from the set leaves the rest of the set to be in the other subset. We exclude $i=0$ and $i=12$ since they leave a set empty. We proceed with complementary counting and casework:

We apply the Principle of Inclusion and Exclusion for casework on the complementary cases. We find the ways where $|A|$ is in $A$, which violates the first condition. Then we find the ways where the elements $|B|$ and $12-|B|$ are in set $B$, which violates only the second condition, and not the first. This ensures we do not overcount.

Case 1: $|A|$ is an element in $A$

There are $\sum_{i=1}^{11}\dbinom{11}{i-1}$ = $2^{11}-1$ ways in this case.

Case 2: $|B|$ and $12-|B|$ are in $B$

We introduce a subcase where $|B|$ is not 6, since the other element would also be 6. There are $B-2$ elements to choose from 10 elements. Therefore, $|B|$ can be from 2 to 11. There are $\sum_{i=2}^{11}\dbinom{10}{i-2}-\dbinom{10}{4} = 2^{10}-211$ ways. The other subcase is when $|B|$ is equal to 6. There are $\dbinom{11}{5}=462$ ways. Adding the subcases gives us $2^{10}+251$.

Adding this with case one gives us $2^{11}+2^{10}+250$. Subtracting this from $2^{12}-2$ gives $1024-252=\boxed{772}$.

~Magnetoninja

See also

2010 AIME II (Problems • Answer Key • Resources)
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Problem 7 		Followed by
Problem 9
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