Difference between revisions of "2019 AMC 10A Problems/Problem 20"
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==Solution 1== | ==Solution 1== | ||
| − | Note that odd sums can only be formed by <math>(e,e,o)</math> or <math>(o,o,o),</math> so we focus on placing the evens to | + | Note that odd sums can only be formed by <math>(e,e,o)</math> or <math>(o,o,o),</math> 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 <math>9</math> ways to do this. There are then <math>5!</math> ways to permute the odd numbers, and <math>4!</math> ways to permute the even numbers, thus giving the answer as <math>\frac{5! \cdot 4! \cdot 9}{9!}=\boxed{\textbf{(B) }\frac{1}{14}}</math>. |
==Solution 2== | ==Solution 2== | ||
| − | By the Pigeonhole Principle, there must be at least one row with <math>2</math> or more odd numbers in it. Therefore, that row must contain <math>3</math> odd numbers in order to have an odd sum. The same thing can be done with the columns. | + | By the Pigeonhole Principle, there must be at least one row with <math>2</math> or more odd numbers in it. Therefore, that row must contain <math>3</math> 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 <math>3 \cdot 3 = 9</math>. The denominator will be <math>\binom{9}{4}</math>, the total number of ways we could choose which <math>4</math> of the <math>9</math> squares will contain an even number. Hence the answer is <cmath>\frac{9}{\binom{9}{4}}=\boxed{\textbf{(B) }\frac{1}{14}}</cmath> |
| − | <cmath>\frac{9}{\binom{9}{4}}=\boxed{\textbf{(B) }\frac{1}{14}}</cmath> | ||
==Video Solution== | ==Video Solution== | ||
Revision as of 00:33, 27 February 2019
- The following problem is from both the 2019 AMC 10A #20 and 2019 AMC 12A #16, so both problems redirect to this page.
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?
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 
.
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 ![\[\frac{9}{\binom{9}{4}}=\boxed{\textbf{(B) }\frac{1}{14}}\]](http://latex.artofproblemsolving.com/b/0/2/b02fd305ab2b24c4b7780359c414686a3db9602a.png)
Video Solution
For those who want a video solution: https://www.youtube.com/watch?v=uJgS-q3-1JE
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. 
