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

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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.
 
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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{{incomplete|solution}}
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==Solution 2==
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Let the two disjoint subsets be <math>A</math> and <math>B</math>, and let <math>C = S-(A+B)</math>.
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For each <math>i \in 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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However, there are <math>2^{10}</math> ways to organize the elements of <math>S</math> such that <math>A = \emptyset</math> and <math>S = B+C</math>, and there are <math>2^{10}</math> ways to organize the elements of <math>S</math> such that <math>B = \emptyset</math> and <math>S = A+C</math>.
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But, the combination such that <math>A = B = \emptyset</math> and <math>S = C</math> is counted twice.
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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>. 
  
 
== 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}}

Revision as of 14:24, 24 June 2008

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

For simplicity, let's call the sets $\mathcal{A}$ and $\mathcal{B}$. Now if we choose $x$ members from $\mathcal{S}$ to be in $\mathcal{A}$, then we have $10-x$ elements to choose for $\mathcal{B}$. Thus

$n=\sum_{x=1}^{9} (\binom{10}{x}\cdot(\sum_{n=1}^{10-x}\binom{10-x}{n}))=\sum_{x=1}^{9} (\binom{10}{x}\cdot(2^{10-x}-1))=\binom{10}{1}\cdot511+\binom{10}{2}\cdot255+\binom{10}{3}\cdot127+\binom{10}{4}\cdot63+\binom{10}{5}\cdot31+\binom{10}{6}\cdot15+\binom{10}{7}\cdot7+\binom{10}{8}\cdot3+\binom{10}{9}\cdot1=\binom{10}{1}\cdot512+\binom{10}{2}\cdot258+\binom{10}{3}\cdot134+\binom{10}{4}\cdot78+\binom{10}{5}\cdot31$.

We want the remainder when $n$ is divided by 1000, so we find the last three digits of each.

Solution 2

Let the two disjoint subsets be $A$ and $B$, and let $C = S-(A+B)$. For each $i \in S$, 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 $S$ such that $A = \emptyset$ and $S = B+C$, and there are $2^{10}$ ways to organize the elements of $S$ such that $B = \emptyset$ and $S = A+C$. But, the combination such that $A = B = \emptyset$ and $S = C$ 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 $\{ A,B \}$, $n = \frac{1}{2}(3^{10}-2\cdot2^{10}+1) = 28501 \equiv \boxed{501} \pmod{1000}$.

See also

2002 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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