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's say that the first set has $x$ elements and the second set has $y$ elements, where $x+y\leq 10$ . The number of $\mathcal{S}$ is $\binom{10}{x}\binom{10-x}{y}=\dfrac{10!(10-x)!}{x!(10-x)!y!(10-x-y)!}=\dfrac{10!}{x!y!(10-x-y)!}$

We now divide by 2, since order doesn't matter in $\mathcal{S}$ : $\dfrac{5\cdot 9!}{x!y!(10-x-y)!}$ . Thus

$n=\sum_{x+y\leq 10} \dfrac{5\cdot 9!}{x!y!(10-x-y)!}.$

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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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