Difference between revisions of "2019 AMC 10A Problems/Problem 20"

Revision as of 20:38, 17 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 $1,2,\dots,9$ are randomly placed into the $9$ squares of a $3 \times 3$ 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?

$\textbf{(A) }\frac{1}{21}\qquad\textbf{(B) }\frac{1}{14}\qquad\textbf{(C) }\frac{5}{63}\qquad\textbf{(D) }\frac{2}{21}\qquad\textbf{(E) }\frac{1}{7}$

Solution 1

Note that odd sums can only be formed by $(e,e,o)$ or $(o,o,o),$ so we focus on placing the evens to get $\frac{5! \cdot 4! \cdot 9}{9!}=\boxed{\textbf{(B) }\frac{1}{14}}$ (we need to have each even be with another even in each row/column).

Solution 2

By the Pigeonhole Principle, there must be at least one row with $2$ or more odd numbers in it. Therefore, that row must contain $3$ odd numbers in order to have an odd sum. The same thing can be done with the columns. Then we simply have to choose one row and one column to be filled with odd numbers - which can happen $3\cdot 3 = 9$ times. The answer is then $\frac{9}{\binom{9}{4}}=\boxed{\textbf{(B) }\frac{1}{14}}$

Video Solution

For those who want a video solution: https://www.youtube.com/watch?v=uJgS-q3-1JE

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.

@@ Line 7: / Line 7: @@
 <math>\textbf{(A) }\frac{1}{21}\qquad\textbf{(B) }\frac{1}{14}\qquad\textbf{(C) }\frac{5}{63}\qquad\textbf{(D) }\frac{2}{21}\qquad\textbf{(E) }\frac{1}{7}</math>
-==Solution==
+==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 get <math>\frac{5! \cdot 4! \cdot 9}{9!}=\boxed{B) \frac{1}{14}}</math> (we need to have each even be with another even in each row/column)
+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 get <math>\frac{5! \cdot 4! \cdot 9}{9!}=\boxed{\textbf{(B) }\frac{1}{14}}</math> (we need to have each even be with another even in each row/column).
 ==Solution 2==
-By the Pigeonhole Principle, there must be at least one row with 2 or more odd numbers in it. Therefore, that row must contain three odd numbers in order to sum to odd. The same thing can be done with the columns. Then we simply have to choose one row and one column to be filled with odd numbers - which can happen <math>3\cdot 3 = 9</math> times. The answer is then
+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. Then we simply have to choose one row and one column to be filled with odd numbers - which can happen <math>3\cdot 3 = 9</math> times. The answer is then
-<cmath>\frac{9}{\binom{9}{4}}=\boxed{\text{(B) }\tfrac{1}{14}}</cmath>
+<cmath>\frac{9}{\binom{9}{4}}=\boxed{\textbf{(B) }\frac{1}{14}}</cmath>
--programjames1
 ==Video Solution==
 For those who want a video solution: https://www.youtube.com/watch?v=uJgS-q3-1JE

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