Difference between revisions of "1998 AIME Problems/Problem 4"
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=== Solution 2 === | === Solution 2 === | ||
Let <math>O</math> stand for an odd number and <math>E</math> an even. | 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. Let's do the case <math>OOO</math>, <math>OEE</math>, <math>OEE</math>. | + | 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{ | + | <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 | - 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 02:37, 12 December 2022
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
Solution 3
For this problem, let's think about parity. There are odd numbers from
and there are four even numbers from
. 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 or
where the letters
and
stand for odd and even respectively.
Since we constrained to only odds, the only way to do the pairing is
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:
Now since we are choosing in groups, we can ignor order of these terms.
For the first , there are
ways to choose which odd to use, and
or
ways to choose the evens.
Great, let's move on to the second . There are
odds left, so there are
ways choose the odds. Since there are only two even numbers, they have to go here, so there is only
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 or
.
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 options and the second player has
. When we multiply these, we get
. This is our denominator.
When we make the fraction, we have . When we simplify it, we have:
We are asked to find the sum of the numerator and the denominator, so summing these, we have:
-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 |
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