Difference between revisions of "2002 AIME II Problems/Problem 9"
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− | Let the two disjoint subsets be <math>A</math> and <math>B</math>, and let <math>C = 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>. | + | 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>S</math> such that <math>A = \emptyset</math> and <math>S = B+C</math>, and there are <math>2^{10}</math> ways to organize the elements of <math>S</math> such that <math>B = \emptyset</math> and <math>S = A+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>S = C</math> is counted twice. | + | 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>. | 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>. |
Latest revision as of 22:07, 4 May 2024
Contents
[hide]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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.