Difference between revisions of "1998 AIME Problems/Problem 4"
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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. | + | 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 00:26, 4 June 2014
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
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 
odd tiles and 
even tiles, the only possibility is that one player gets 
odd tiles and the other two players get 
even tiles and 
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 choices for the player who will get three odd tiles, and 
choices for the tiles that he gets. The other two odd tiles can be distributed to the other two players in ways, and the even tiles can be distributed between them in 
ways. This gives us a total of 
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 
ways, and the second player needs three of the remaining six, which we can give him in 
ways. Finally, the third player will simply take the remaining tiles in way. So, there are 
ways total to distribute the tiles.
Thus, the total probability is 
so the answer is 
.
See also
| 1998 AIME (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
