1998 AIME Problems/Problem 4
Contents
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, he must have an odd number of odd tiles: that is, he 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.)
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.
We must multiply the probability by 3, since any of the 3 players can have the 3 odd tiles.Thus, the total probability is so the answer is .
Solution 2
Let stand for an odd number and an even. Therefore, one person must pick , the other person must pick and the last person must pick . Since any permutation of the order of who is picking or change in the order of the even numbers (e.g. instead of ) doesn't change the probability, we just need to multiply the probability of one case by as there are 27 such cases (by cases I mean ordered triples of ordered multisets such that one of them has 3 's and the other two have two 's and an in them, respectively.) . Let's do the case , , . . We now multiply by 27 to get - whatRthose
See also
1998 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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