Difference between revisions of "1998 AIME Problems/Problem 4"
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== Problem == | == Problem == | ||
| − | 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 | + | 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 integer]]s. Find <math>m+n.</math> |
== Solution == | == Solution == | ||
| − | {{ | + | There are 5 odd tiles and 4 [[even]] tiles. In order for a player to have an odd sum, she must have an odd number of odd tiles. Thus, the only possibility is that one player gets three odd tiles and the other two players get two even and one odd tile. We count the number of ways this can happen: |
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| + | We have 3 choices for the player who can get all odd tiles, and <math>{5 \choose 3} = 10</math> choices for the tiles that he gets. The other two odd tiles can be distributed to the other two players in 2 ways, and the even tiles can be distributed between them in <math>{4 \choose 2} = 6</math> ways. This gives us a total of <math>3\cdot 10 \cdot 2 \cdot 6 = 360</math> possibilities. | ||
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| + | In order to calculate the probability, we need to know the total number of possible distributions for the tiles. The first player needs three tiles which we can give him in <math>{9 \choose 3} = 84</math> ways, and the second player needs three of the remaining six, which we can give him in <math>{6 \choose 3} = 20</math> ways. | ||
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| + | Thus, the total probability is <math>\frac{360}{84 \cdot 20} = \frac{3}{14}</math> so the answer is <math>3 + 14 = 017</math>. | ||
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== See also == | == See also == | ||
* [[1998 AIME Problems/Problem 3 | Previous problem]] | * [[1998 AIME Problems/Problem 3 | Previous problem]] | ||
* [[1998 AIME Problems/Problem 5 | Next problem]] | * [[1998 AIME Problems/Problem 5 | Next problem]] | ||
* [[1998 AIME Problems]] | * [[1998 AIME Problems]] | ||
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| + | [[Category:Intermediate Combinatorics Problems]] | ||
Revision as of 11:26, 29 January 2007
Problem
Nine tiles are numbered 
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 
where 
and 
are relatively prime positive integers. Find 
Solution
There are 5 odd tiles and 4 even tiles. In order for a player to have an odd sum, she must have an odd number of odd tiles. Thus, the only possibility is that one player gets three odd tiles and the other two players get two even and one odd tile. We count the number of ways this can happen:
We have 3 choices for the player who can get all odd tiles, and 
choices for the tiles that he gets. The other two odd tiles can be distributed to the other two players in 2 ways, and the even tiles can be distributed between them in 
ways. This gives us a total of 
possibilities.
In order to calculate the probability, we need to know the total number of possible distributions for the tiles. The first player needs three tiles which we can give him in 
ways, and the second player needs three of the remaining six, which we can give him in 
ways.
Thus, the total probability is 
so the answer is 
.
