2019 AMC 10A Problems/Problem 20

Revision as of 18:28, 9 February 2019 by Programjames1 (talk | contribs) (Solution)
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

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 $\frac{5! \cdot 4! \cdot 9}{9!}=\boxed{B) \frac{1}{14}}$ (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 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 $3\cdot 3 = 9$ times. The answer is then \[\frac{9}{\binom{9}{4}}=\boxed{\text{(B) }\tfrac{1}{14}}\]

-programjames1

See Also

2019 AMC 10A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

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