Difference between revisions of "2019 AMC 10A Problems/Problem 20"
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Therefore our answer is simply <cmath>\frac{5! \cdot 4! \cdot 9}{9!} = \frac{1}{7\cdot 2} = \boxed{\frac{1}{14}}.</cmath> | Therefore our answer is simply <cmath>\frac{5! \cdot 4! \cdot 9}{9!} = \frac{1}{7\cdot 2} = \boxed{\frac{1}{14}}.</cmath> | ||
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== Video Solutions == | == Video Solutions == |
Revision as of 15:06, 17 June 2021
- The following problem is from both the 2019 AMC 10A #20 and 2019 AMC 12A #16, so both problems redirect to this page.
Contents
Problem
The numbers are randomly placed into the squares of a grid. Each square gets one number, and each of the numbers is used once. What is the probability that the sum of the numbers in each row and each column is odd?
Solutions
Solution 1
Note that odd sums can only be formed by or so we focus on placing the evens: we need to have each even be with another even in each row/column. It can be seen that there are ways to do this. There are then ways to permute the odd numbers, and ways to permute the even numbers, thus giving the answer as .
~Petallstorm
Solution 2
By the Pigeonhole Principle, there must be at least one row with or more odd numbers in it. Therefore, that row must contain odd numbers in order to have an odd sum. The same thing can be done with the columns. Thus we simply have to choose one row and one column to be filled with odd numbers, so the number of valid odd/even configurations (without regard to which particular odd and even numbers are placed where) is . The denominator will be , the total number of ways we could choose which of the squares will contain an even number. Hence the answer is
- The Pigeonhole Principle isn't really necessary here: After noting from the first solution that any row that contains evens must contain two evens, the result follows that the four evens must form the corners of a rectangle.
~Petallstorm
Solution 3
Note that there are 5 odds and 4 evens, and for three numbers to sum an odd number, either 1 or three must be odd. Hence, one column must be all odd and one row must be all odd. First, we choose a row, for which there are three choices and within the row P(5,3). There are 3! ways to order the remaining odds and 4! ways to order the evens. The total possible ways is 9!.
~Petallstorm
Solution 4
Note that the odd sums are only formed by, or any permutation of . When looking at a 3x3 box, we realize that there must always be one column that is odd and one row that is odd. Calculating the probability of one permutation of this we get:
Odd - | Odd - | Odd - |
Odd - | Even | Even |
Odd - | Even | Even |
Now, there are 9 ways you can get this particular permutation (3 choices for all odd row, 3 choices for all odd column), so multiplying the result by , we get:
~petallstorm
Solution 5
To get an odd sum we need or we need . Note that there are ways to arrange the numbers. Also notice that there are even numbers and odd numbers from through .
The number of ways to arrange the even numbers is and the number of ways to arrange the odd numbers is . Now a common strategy is to deal with the category with fewer numbers so this suggests for us to look at even numbers and try something out.
Since we have even numbers we need to split two evens to one row and two evens to another row. Note that there are ways to choose the rows, and the to choose where the two even numbers go in that row. The odd numbers go wherever there is a spot left which means we don't need to care about them. So we totally have ways to place the even numbers.
Don't forget, we still need to arrange these even and odd numbers and like mentioned above there are and ways respectively.
Therefore our answer is simply
Video Solutions
Video Solution 1
Education, the Study of Everything
Video Solution 2
https://www.youtube.com/watch?v=uJgS-q3-1JE
Video Solution 3
https://www.youtube.com/watch?v=3Sj7XPernCY
Video Solution 4
https://www.youtube.com/watch?v=APKGHtj-2rI
Video Solution 5
https://youtu.be/IRyWOZQMTV8?t=2069
~ pi_is_3.14
See Also
2019 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
2019 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 15 |
Followed by Problem 17 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.