Difference between revisions of "1998 AIME Problems/Problem 4"

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== Problem ==
 
== Problem ==
Nine tiles are numbered <math>1, 2, 3, \ldots, 9,</math> respectively.  Each of three players randomly selects and keeps three of the tile, and sums those three values.  The probability that all three players obtain an odd sum is <math>m/n,</math> where <math>m</math> and <math>n</math> are relatively prime positive integers.  Find <math>m+n.</math>
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Nine tiles are numbered <math>1, 2, 3, \cdots, 9,</math> respectively.  Each of three players randomly selects and keeps three of the tiles, and sums those three values.  The probability that all three players obtain an odd sum is <math>m/n,</math> where <math>m</math> and <math>n</math> are relatively prime positive integers.  Find <math>m+n.</math>
  
 
== Solution ==
 
== Solution ==

Revision as of 08:40, 27 January 2007

Problem

Nine tiles are numbered $1, 2, 3, \cdots, 9,$ respectively. Each of three players randomly selects and keeps three of the tiles, and sums those three values. The probability that all three players obtain an odd sum is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution

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See also