Difference between revisions of "2002 AIME II Problems/Problem 9"
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We want the remainder when <math>n</math> is divided by 1000, so we find the last three digits of each. | We want the remainder when <math>n</math> is divided by 1000, so we find the last three digits of each. | ||
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==Solution 2== | ==Solution 2== |
Revision as of 13:25, 24 June 2008
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
For simplicity, let's call the sets and . Now if we choose members from to be in , then we have elements to choose for . Thus
.
We want the remainder when is divided by 1000, so we find the last three digits of each.
Solution 2
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 , .
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 |