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Difference between revisions of "2002 AIME II Problems/Problem 9"
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== Solution 1 ==
 
== Solution 1 ==
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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>. 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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Let the two disjoint subsets be <math>A</math> and <math>B</math>, and let <math>C = 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>.  
  
 
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>.  
 
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>.  

Revision as of 22:06, 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 $C = S-(A+B)$. 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 $S$ 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 $\{ A,B \}$, $n = \frac{1}{2}(3^{10}-2\cdot2^{10}+1) = 28501 \equiv \boxed{501} \pmod{1000}$.

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
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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