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

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

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

Page

Toolbox

Search

2002 AIME II Problems/Problem 9

Problem

Solution

See also