Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

1998 AIME Problems/Problem 4

Revision as of 14:46, 10 June 2019 by Whatrthose (talk | contribs) (→‎Solution)

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, 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 $5$ 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.)

$\dbinom{5}{3} = 10$ 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 $10 \cdot 2 \cdot 6 = 120$ 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.

We must multiply the probability by 3, since any of the 3 players can have the 3 odd tiles.Thus, the total probability is $\frac{360}{1680} = \frac{3}{14},$ so the answer is $3 + 14 = \boxed{017}$.

Solution 2

Let $O$ stand for an odd number and $E$ an even. Therefore, one person must pick $OOO$, the other person must pick $OEE$ and the last person must pick $OOO$. Since any permutation of the order of who is picking or change in the order of the even numbers (e.g. $EOE$ instead of $OEE$) doesn't change the probability, we just need to multiply the probability of one case by $\binom{3}{2}^3 = 27$ as there are 27 such cases. Let's do the case $OOO$, $OEE$, $OEE$. $\frac{5}{9} \times \frac{4}{8} \times \frac{3}{7} \times \frac{2}{6} \times \frac{4}{5} \times \frac{3}{4} \times \frac{1}{3} \times \frac{2}{2} \times \frac{1}{1} = \frac{1}{126}$. We now multiply by 27 to get $\frac{3}{14} \implies m + n = \boxed{17}$

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1998_AIME_Problems/Problem_4&oldid=106237"