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Difference between revisions of "2002 AIME II Problems/Problem 9"

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== Problem ==
 
== Problem ==
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Let <math>\mathcal{S}</math> be the set <math>\lbrace1,2,3,\ldots,10\rbrace</math> Let <math>n</math> be the number of sets of two non-empty disjoint subsets of <math>\mathcal{S}</math>. (Disjoint sets are defined as sets that have no common elements.) Find the remainder obtained when <math>n</math> is divided by <math>1000</math>.
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Let <math>\mathcal{S}</math> be the [[set]] <math>\lbrace1,2,3,\ldots,10\rbrace</math> Let <math>n</math> be the number of sets of two non-empty disjoint subsets of <math>\mathcal{S}</math>. (Disjoint sets are defined as sets that have no common elements.) Find the remainder obtained when <math>n</math> is divided by <math>1000</math>.
  
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== Solution ==
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== Solution 1 ==
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For simplicity, let's call the sets <math>\mathcal{A}</math> and <math>\mathcal{B}</math>. Now if we choose <math>x</math> members from <math>\mathcal{S}</math> to be in <math>\mathcal{A}</math>, then we have <math>10-x</math> elements to choose for <math>\mathcal{B}</math>. Thus
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Let the two disjoint subsets be <math>A</math> and <math>B</math>, and let <math>C = \mathcal{S}-(A+B)</math>. For each <math>i \in \mathcal{S}</math>, either <math>i \in A</math>, <math>i \in B</math>, or <math>i \in C</math>. So there are <math>3^{10}</math> ways to organize the elements of <math>S</math> into disjoint <math>A</math>, <math>B</math>, and <math>C</math>.  
  
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<math>n=\sum_{x=1}^{9} (\binom{10}{x}\cdot(\sum_{n=1}^{10-x}\binom{10-x}{n}))=\sum_{x=1}^{9} (\binom{10}{x}\cdot(2^{10-x}-1))=\binom{10}{1}\cdot511+\binom{10}{2}\cdot255+\binom{10}{3}\cdot127+\binom{10}{4}\cdot63+\binom{10}{5}\cdot31+\binom{10}{6}\cdot15+\binom{10}{7}\cdot7+\binom{10}{8}\cdot3+\binom{10}{9}\cdot1=\binom{10}{1}\cdot512+\binom{10}{2}\cdot258+\binom{10}{3}\cdot134+\binom{10}{4}\cdot78+\binom{10}{5}\cdot31</math>.
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However, there are <math>2^{10}</math> ways to organize the elements of <math>\mathcal{S}</math> such that <math>A = \emptyset</math> and <math>\mathcal{S} = B+C</math>, and there are <math>2^{10}</math> ways to organize the elements of <math>\mathcal{S}</math> such that <math>B = \emptyset</math> and <math>\mathcal{S} = A+C</math>.
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But, the combination such that <math>A = B = \emptyset</math> and <math>\mathcal{S} = C</math> is counted twice.  
  
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We want the remainder when <math>n</math> is divided by 1000, so we find the last three digits of each.
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Thus, there are <math>3^{10}-2\cdot2^{10}+1</math> ordered pairs of sets <math>(A,B)</math>. But since the question asks for the number of unordered sets <math>\{ A,B \}</math>, <math>n = \frac{1}{2}(3^{10}-2\cdot2^{10}+1) = 28501 \equiv \boxed{501} \pmod{1000}</math>.
  
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{{incomplete|solution}}
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== Solution 2 ==
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Let <math>A</math> and <math>B</math> be the disjoint subsets. If <math>A</math> has <math>n</math> elements, then the number of elements of <math>B</math> can be any positive integer number less than or equal to <math>10-n</math>. So <math>2n=\binom{10}{1} \cdot \left(\binom{9}{1}+\binom{9}{2}+\dots +\binom{9}{9}\right)+\binom{10}{2} \cdot \left(\binom{8}{1}+\binom{8}{2}+\dots +\binom{8}{8}\right)+\dots +\binom{10}{9} \cdot \binom{1}{1}=</math>
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 +
<math>=\binom{10}{1} \cdot \sum_{n=1}^9 \binom{9}{n}+\binom{10}{2} \cdot \sum_{n=1}^8 \binom{8}{n}+\dots + \binom{10}{9} \cdot \binom{1}{1}=</math>
 +
 
 +
<math>=\binom{10}{1} \cdot \left(2^9-1\right)+\binom{10}{2} \cdot \left(2^8-1\right)+\dots+\binom{10}{9} \cdot \left(2-1\right)=</math>
 +
 
 +
<math>=\sum_{n=0}^{10} \binom{10}{n} 2^{10-n} 1^n - \binom{10}{0} \cdot 2^{10} - \binom{10}{10}-\left(\sum_{n=0}^{10} \binom{10}{n} - \binom{10}{0} - \binom{10}{10} \right) =</math>
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<math>=(2+1)^{10}-2^{10}-1-(1+1)^{10}+1+1=3^{10}-2^{11}+1=57002</math>
 +
 
 +
Then <math>n=\frac{57002}{2}=28501\equiv \boxed{501} \pmod{1000}</math>
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2002|n=II|num-b=8|num-a=10}}
 
{{AIME box|year=2002|n=II|num-b=8|num-a=10}}
 +
 +
[[Category:Intermediate Combinatorics Problems]]
 +
{{MAA Notice}}

Latest revision as of 22:07, 4 May 2024

Problem

Let $\mathcal{S}$ be the set $\lbrace1,2,3,\ldots,10\rbrace$ Let $n$ be the number of sets of two non-empty disjoint subsets of $\mathcal{S}$. (Disjoint sets are defined as sets that have no common elements.) Find the remainder obtained when $n$ is divided by $1000$.

Solution 1

Let the two disjoint subsets be $A$ and $B$, and let $C = \mathcal{S}-(A+B)$. For each $i \in \mathcal{S}$, either $i \in A$, $i \in B$, or $i \in C$. So there are $3^{10}$ ways to organize the elements of $S$ into disjoint $A$, $B$, and $C$.

However, there are $2^{10}$ ways to organize the elements of $\mathcal{S}$ such that $A = \emptyset$ and $\mathcal{S} = B+C$, and there are $2^{10}$ ways to organize the elements of $\mathcal{S}$ such that $B = \emptyset$ and $\mathcal{S} = A+C$. But, the combination such that $A = B = \emptyset$ and $\mathcal{S} = C$ is counted twice.

Thus, there are $3^{10}-2\cdot2^{10}+1$ ordered pairs of sets $(A,B)$. But since the question asks for the number of unordered sets $\{ A,B \}$, $n = \frac{1}{2}(3^{10}-2\cdot2^{10}+1) = 28501 \equiv \boxed{501} \pmod{1000}$.

Solution 2

Let $A$ and $B$ be the disjoint subsets. If $A$ has $n$ elements, then the number of elements of $B$ can be any positive integer number less than or equal to $10-n$. So $2n=\binom{10}{1} \cdot \left(\binom{9}{1}+\binom{9}{2}+\dots +\binom{9}{9}\right)+\binom{10}{2} \cdot \left(\binom{8}{1}+\binom{8}{2}+\dots +\binom{8}{8}\right)+\dots +\binom{10}{9} \cdot \binom{1}{1}=$

$=\binom{10}{1} \cdot \sum_{n=1}^9 \binom{9}{n}+\binom{10}{2} \cdot \sum_{n=1}^8 \binom{8}{n}+\dots + \binom{10}{9} \cdot \binom{1}{1}=$

$=\binom{10}{1} \cdot \left(2^9-1\right)+\binom{10}{2} \cdot \left(2^8-1\right)+\dots+\binom{10}{9} \cdot \left(2-1\right)=$

$=\sum_{n=0}^{10} \binom{10}{n} 2^{10-n} 1^n - \binom{10}{0} \cdot 2^{10} - \binom{10}{10}-\left(\sum_{n=0}^{10} \binom{10}{n} - \binom{10}{0} - \binom{10}{10} \right) =$

$=(2+1)^{10}-2^{10}-1-(1+1)^{10}+1+1=3^{10}-2^{11}+1=57002$

Then $n=\frac{57002}{2}=28501\equiv \boxed{501} \pmod{1000}$

See also

2002 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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