2002 AIME II Problems/Problem 9

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.

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2002 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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2002 AIME II Problems/Problem 9

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