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

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== 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. (Below, we say that the people are distinguishable - say that they are named Ana, Beth, and Clark, for example.)
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In order for a player to have an odd sum, she must have an odd number of odd tiles: that is, she can either have three odd tiles, or two even tiles and an odd tile.  Thus, since there are <math>5</math> odd tiles and <math>4</math> even tiles, the only possibility is that one player gets <math>3</math> odd tiles and the other two players get <math>2</math> even tiles and <math>1</math> odd tile.  We count the number of ways this can happen. (We will count assuming that it matters in what order the people pick the tiles; the final answer is the same if we assume the opposite, that order doesn't matter.)
  
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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We have <math>3</math> 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 <math>2</math> 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.
  
 
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.
 
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.

Revision as of 01:26, 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

In order for a player to have an odd sum, she must have an odd number of odd tiles: that is, she can either have three odd tiles, or two even tiles and an odd tile. Thus, since there are $5$ odd tiles and $4$ even tiles, the only possibility is that one player gets $3$ odd tiles and the other two players get $2$ even tiles and $1$ odd tile. We count the number of ways this can happen. (We will count assuming that it matters in what order the people pick the tiles; the final answer is the same if we assume the opposite, that order doesn't matter.)

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

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