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Difference between revisions of "1998 AIME Problems/Problem 4"
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== Solution ==
 
== Solution ==
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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:
+
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. (Below, we say that the people are distinguishable - say that they are named Ana, Beth, and Clark, for example.)
  
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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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We have 3 choices for the player who will get three odd tiles, and <math>\dbinom{5}{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>\dbinom{4}{2} \cdot \dbinom{2}{2} = 6</math> ways.  This gives us a total of <math>3\cdot 10 \cdot 2 \cdot 6 = 360</math> possibilities in which all three people get odd sums.
  
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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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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>\dbinom{9}{3} = 84</math> ways, and the second player needs three of the remaining six, which we can give him in <math>\dbinom{6}{3} = 20</math> ways. Finally, the third player will simply take the remaining tiles in <math>1</math> way. So, there are <math>\dbinom{9}{3} \cdot \dbinom{6}{3} \cdot 1 = 84 \cdot 20 = 1680</math> ways total to distribute the tiles.
  
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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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Thus, the total probability is <math>\frac{360}{1680} = \frac{3}{14},</math> so the answer is <math>3 + 14 = \boxed{017}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 00:21, 4 June 2014

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

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. (Below, we say that the people are distinguishable - say that they are named Ana, Beth, and Clark, for example.)

We have 3 choices for the player who will get three odd tiles, and $\dbinom{5}{3} = 10$ 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 $\dbinom{4}{2} \cdot \dbinom{2}{2} = 6$ ways. This gives us a total of $3\cdot 10 \cdot 2 \cdot 6 = 360$ possibilities in which all three people get odd sums.

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 $\dbinom{9}{3} = 84$ ways, and the second player needs three of the remaining six, which we can give him in $\dbinom{6}{3} = 20$ ways. Finally, the third player will simply take the remaining tiles in $1$ way. So, there are $\dbinom{9}{3} \cdot \dbinom{6}{3} \cdot 1 = 84 \cdot 20 = 1680$ ways total to distribute the tiles.

Thus, the total probability is $\frac{360}{1680} = \frac{3}{14},$ so the answer is $3 + 14 = \boxed{017}$.

See also

1998 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
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All AIME Problems and Solutions

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