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

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

(→‎Solution)
(→‎Solution 2)
 
(13 intermediate revisions by 5 users not shown)
Line 3: Line 3:
  
 
== Solution ==
 
== 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 <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.)
+
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 <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 <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.
+
<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>10 \cdot 2 \cdot 6 = 120</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.
  
−
Thus, the total probability is <math>\frac{360}{1680} = \frac{3}{14},</math> so the answer is <math>3 + 14 = \boxed{017}</math>.
+
We must multiply the probability by 3, since any of the 3 players can have the 3 odd tiles.Thus, the total probability is <math>\frac{360}{1680} = \frac{3}{14},</math> so the answer is <math>3 + 14 = \boxed{017}</math>.
 +
=== Solution 2 ===
 +
Let <math>O</math> stand for an odd number and <math>E</math> an even.
 +
Therefore, one person must pick <math>OOO</math>, the other person must pick <math>OEE</math> and the last person must pick <math>OEE</math>. Since any permutation of the order of who is picking or change in the order of the even numbers (e.g. <math>EOE</math> instead of <math>OEE</math>) doesn't change the probability, we just need to multiply the probability of one case by <math>\binom{3}{2}^3 = 27</math> as there are 27 such cases (by cases I mean ordered triples of ordered multisets <math>A, B, C</math> such that one of them has 3 <math>O</math>'s and the other two have two <math>E</math>'s and an <math>O</math> in them, respectively.) . Let's do the case <math>OOO</math>, <math>OEE</math>, <math>OEE</math>.
 +
<math>\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}</math>. We now multiply by 27 to get <math>\frac{3}{14} \implies m + n = \boxed{017}</math>
 +
- whatRthose
 +
 
 +
=== Solution 3 ===
 +
 
 +
For this problem, let's think about parity. There are <math>5</math> odd numbers from <math>1-9</math> and there are four even numbers from <math>1-9</math>. Since this problem is asking for the probability that the each player gets an odd sum, we also have to calculate the total numeber of ways.
 +
 
 +
In this case, there are only two ways to get an odd sum. Either have the sequence <math>OOO</math> or <math>OEE</math> where the letters <math>O</math> and <math>E</math> stand for odd and even respectively.
 +
 
 +
Since we constrained to only <math>5</math> odds, the only way to do the pairing is
 +
<cmath>OOO-OEE-OEE</cmath>
 +
There are of couse three ways to choose who gets the three odds.
 +
 
 +
Once we have chosen who has gotten the three odds, we can actually reorder the sequence like this:
 +
<cmath>OEE-OEE-OOO</cmath>
 +
Now since we are choosing in groups, we can ignor order of these terms.
 +
 
 +
For the first <math>OEE</math>, there are <math>5</math> ways to choose which odd to use, and <math>\dbinom{4}{2}</math> or <math>6</math> ways to choose the evens.
 +
 
 +
Great, let's move on to the second <math>OEE</math>. There are <math>4</math> odds left, so there are <math>4</math> ways choose the odds. Since there are only two even numbers, they have to go here, so there is only <math>1</math> way to choose the evens.
 +
 
 +
The three odds will now fall in place. We can now multiply all the numbers since they are independent, and we have <math>5*6*4*3</math> or <math>360</math>.
 +
 
 +
Since this is a probability question, we have to ask ourselves how many ways are there to distribute the 9 tiles equally among 3 players.
 +
 
 +
Fortunately for us, this is not hard as the first player has <math>\dbinom{9}{3}</math> options and the second player has <math>\dbinom{6}{3}</math>. When we multiply these, we get <math>1680</math>. This is our denominator.
 +
 
 +
When we make the fraction, we have <math>\frac{360}{1680}</math>. When we simplify it, we have:
 +
<cmath>\frac{3}{14}</cmath>
 +
We are asked to find the sum of the numerator and the denominator, so summing these, we have:
 +
<cmath>\boxed{017}</cmath>
 +
-Pi_is_3.14
  
 
== See also ==
 
== See also ==

Latest revision as of 01:37, 12 December 2022

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 $OEE$. 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 (by cases I mean ordered triples of ordered multisets $A, B, C$ such that one of them has 3 $O$'s and the other two have two $E$'s and an $O$ in them, respectively.) . 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 $\frac{3}{14} \implies m + n = \boxed{017}$ - whatRthose

Solution 3

For this problem, let's think about parity. There are $5$ odd numbers from $1-9$ and there are four even numbers from $1-9$. Since this problem is asking for the probability that the each player gets an odd sum, we also have to calculate the total numeber of ways.

In this case, there are only two ways to get an odd sum. Either have the sequence $OOO$ or $OEE$ where the letters $O$ and $E$ stand for odd and even respectively.

Since we constrained to only $5$ odds, the only way to do the pairing is \[OOO-OEE-OEE\] There are of couse three ways to choose who gets the three odds.

Once we have chosen who has gotten the three odds, we can actually reorder the sequence like this: \[OEE-OEE-OOO\] Now since we are choosing in groups, we can ignor order of these terms.

For the first $OEE$, there are $5$ ways to choose which odd to use, and $\dbinom{4}{2}$ or $6$ ways to choose the evens.

Great, let's move on to the second $OEE$. There are $4$ odds left, so there are $4$ ways choose the odds. Since there are only two even numbers, they have to go here, so there is only $1$ way to choose the evens.

The three odds will now fall in place. We can now multiply all the numbers since they are independent, and we have $5*6*4*3$ or $360$.

Since this is a probability question, we have to ask ourselves how many ways are there to distribute the 9 tiles equally among 3 players.

Fortunately for us, this is not hard as the first player has $\dbinom{9}{3}$ options and the second player has $\dbinom{6}{3}$. When we multiply these, we get $1680$. This is our denominator.

When we make the fraction, we have $\frac{360}{1680}$. When we simplify it, we have: \[\frac{3}{14}\] We are asked to find the sum of the numerator and the denominator, so summing these, we have: \[\boxed{017}\] -Pi_is_3.14

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=183885"