Difference between revisions of "2002 AIME II Problems/Problem 9"
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== Problem == | == Problem == | ||
+ | Let <math>\mathcal{S}</math> be the [[set]] <math>\lbrace1,2,3,\ldots,10\rbrace</math> Let <math>n</math> be the number of sets of two non-empty disjoint subsets of <math>\mathcal{S}</math>. (Disjoint sets are defined as sets that have no common elements.) Find the remainder obtained when <math>n</math> is divided by <math>1000</math>. | ||
− | == Solution == | + | == Solution 1 == |
+ | Let the two disjoint subsets be <math>A</math> and <math>B</math>, and let <math>C = \mathcal{S}-(A+B)</math>. For each <math>i \in \mathcal{S}</math>, either <math>i \in A</math>, <math>i \in B</math>, or <math>i \in C</math>. So there are <math>3^{10}</math> ways to organize the elements of <math>S</math> into disjoint <math>A</math>, <math>B</math>, and <math>C</math>. | ||
+ | |||
+ | However, there are <math>2^{10}</math> ways to organize the elements of <math>\mathcal{S}</math> such that <math>A = \emptyset</math> and <math>\mathcal{S} = B+C</math>, and there are <math>2^{10}</math> ways to organize the elements of <math>\mathcal{S}</math> such that <math>B = \emptyset</math> and <math>\mathcal{S} = A+C</math>. | ||
+ | But, the combination such that <math>A = B = \emptyset</math> and <math>\mathcal{S} = C</math> is counted twice. | ||
+ | |||
+ | Thus, there are <math>3^{10}-2\cdot2^{10}+1</math> ordered pairs of sets <math>(A,B)</math>. But since the question asks for the number of unordered sets <math>\{ A,B \}</math>, <math>n = \frac{1}{2}(3^{10}-2\cdot2^{10}+1) = 28501 \equiv \boxed{501} \pmod{1000}</math>. | ||
+ | |||
+ | == Solution 2 == | ||
+ | Let <math>A</math> and <math>B</math> be the disjoint subsets. If <math>A</math> has <math>n</math> elements, then the number of elements of <math>B</math> can be any positive integer number less than or equal to <math>10-n</math>. So <math>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}=</math> | ||
+ | |||
+ | <math>=\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}=</math> | ||
+ | |||
+ | <math>=\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)=</math> | ||
+ | |||
+ | <math>=\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) =</math> | ||
+ | |||
+ | <math>=(2+1)^{10}-2^{10}-1-(1+1)^{10}+1+1=3^{10}-2^{11}+1=57002</math> | ||
+ | |||
+ | Then <math>n=\frac{57002}{2}=28501\equiv \boxed{501} \pmod{1000}</math> | ||
== See also == | == See also == | ||
− | + | {{AIME box|year=2002|n=II|num-b=8|num-a=10}} | |
+ | |||
+ | [[Category:Intermediate Combinatorics Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 22:07, 4 May 2024
Contents
Problem
Let be the set Let be the number of sets of two non-empty disjoint subsets of . (Disjoint sets are defined as sets that have no common elements.) Find the remainder obtained when is divided by .
Solution 1
Let the two disjoint subsets be and , and let . For each , either , , or . So there are ways to organize the elements of into disjoint , , and .
However, there are ways to organize the elements of such that and , and there are ways to organize the elements of such that and . But, the combination such that and is counted twice.
Thus, there are ordered pairs of sets . But since the question asks for the number of unordered sets , .
Solution 2
Let and be the disjoint subsets. If has elements, then the number of elements of can be any positive integer number less than or equal to . So
Then
See also
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 |
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