2019 AMC 10 A
Complete problem set with solutions and individual problem pages
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? (2019 AMC 10A Problem, Question#20)
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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 .
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 .
