# 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 1

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 $\{ A,B \}$, $n = \frac{1}{2}(3^{10}-2\cdot2^{10}+1) = 28501 \equiv \boxed{501} \pmod{1000}$.

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