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.

Template:Incomplete

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

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

Contents

Problem

Solution

Solution 2

See also