Difference between revisions of "2002 AIME II Problems/Problem 9"
(It's wrong, plugged in x=8, y=2.) |
(SOMEONE ELSE DO IT.) |
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== Solution == | == Solution == | ||
| − | {{solution}} | + | For simplicity, let's call the sets <math>\mathcal{A}</math> and <math>\mathcal{B}</math>. Now if we choose <math>x</math> members from <math>\mathcal{S}</math> to be in <math>\mathcal{A}</math>, then we have <math>10-x</math> elements to choose for <math>\mathcal{B}</math>. Thus |
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| + | <math>n=\sum_{x=1}^{9} (\binom{10}{x}\cdot(\sum_{n=1}^{10-x}\binom{10-x}{n}))=\sum_{x=1}^{9} (\binom{10}{x}\cdot(2^{10-x}-1))=\binom{10}{1}\cdot511+\binom{10}{2}\cdot255+\binom{10}{3}\cdot127+\binom{10}{4}\cdot63+\binom{10}{5}\cdot31+\binom{10}{6}\cdot15+\binom{10}{7}\cdot7+\binom{10}{8}\cdot3+\binom{10}{9}\cdot1=\binom{10}{1}\cdot512+\binom{10}{2}\cdot258+\binom{10}{3}\cdot134+\binom{10}{4}\cdot78+\binom{10}{5}\cdot31</math>. | ||
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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. | ||
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| + | {{incomplete|solution}} | ||
== See also == | == See also == | ||
{{AIME box|year=2002|n=II|num-b=8|num-a=10}} | {{AIME box|year=2002|n=II|num-b=8|num-a=10}} | ||
Revision as of 10:59, 24 June 2008
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.
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 | ||
