Difference between revisions of "2012 AIME II Problems/Problem 14"

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== Problem 14 ==
 
== Problem 14 ==
 
In a group of nine people each person shakes hands with exactly two of the other people from the group. Let <math>N</math> be the number of ways this handshaking can occur. Consider two handshaking arrangements different if and only if at least two people who shake hands under one arrangement do not shake hands under the other arrangement. Find the remainder when <math>N</math> is divided by <math>1000</math>.
 
In a group of nine people each person shakes hands with exactly two of the other people from the group. Let <math>N</math> be the number of ways this handshaking can occur. Consider two handshaking arrangements different if and only if at least two people who shake hands under one arrangement do not shake hands under the other arrangement. Find the remainder when <math>N</math> is divided by <math>1000</math>.
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== Solution ==
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== See also ==
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{{AIME box|year=2012|n=II|num-b=13|num-a=15}}

Revision as of 17:22, 31 March 2012

Problem 14

In a group of nine people each person shakes hands with exactly two of the other people from the group. Let $N$ be the number of ways this handshaking can occur. Consider two handshaking arrangements different if and only if at least two people who shake hands under one arrangement do not shake hands under the other arrangement. Find the remainder when $N$ is divided by $1000$.


Solution

See also

2012 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions