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 == | ||
| + | {{AIME box|year=2012|n=II|num-b=13|num-a=15}} | ||
Revision as of 16:22, 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 
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 is divided by 
.
Solution
See also
| 2012 AIME II (Problems • Answer Key • Resources) | ||
| 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 | ||
