Difference between revisions of "2002 AIME II Problems/Problem 9"

Revision as of 14:54, 19 April 2008

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

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 == Solution ==
-Let's say that the first set has <math>x</math> elements and the second set has <math>y</math> elements, where <math>x+y\leq 10</math>. The number of <math>\mathcal{S}</math> is <math>\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)!}</math>
-We now divide by 2, since order doesn't matter in <math>\mathcal{S}</math>: <math>\dfrac{5\cdot 9!}{x!y!(10-x-y)!}</math>. Thus
-<cmath>n=\sum_{x+y\leq 10} \dfrac{5\cdot 9!}{x!y!(10-x-y)!}.</cmath>
 {{solution}}
 == See also ==
 {{AIME box|year=2002|n=II|num-b=8|num-a=10}}

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Difference between revisions of "2002 AIME II Problems/Problem 9"

Revision as of 14:54, 19 April 2008

Problem

Solution

See also