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

Revision as of 13:30, 27 October 2020

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: we need to have each even be with another even in each row/column. It can be seen that there are $9$ ways to do this. There are then $5!$ ways to permute the odd numbers, and $4!$ ways to permute the even numbers, thus giving the answer as $\frac{5! \cdot 4! \cdot 9}{9!}=\boxed{\textbf{(B) }\frac{1}{14}}$ .

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. 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 $3 \cdot 3 = 9$ . The denominator will be $\binom{9}{4}$ , the total number of ways we could choose which $4$ of the $9$ squares will contain an even number. Hence the answer is $\frac{9}{\binom{9}{4}}=\boxed{\textbf{(B) }\frac{1}{14}}$

- 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.

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!. $\frac{3 \cdot P(5,3) \cdot 3! \cdot 4!}{9!}=\boxed{\textbf{(B) }\frac{1}{14}}$

-Lingjun

Solution 4

Note that the odd sums are only formed by, $(O, O, O)$ or any permutation of $(O, E, E)$ . 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 - $\frac{5}{9}$	Odd - $\frac{4}{8}$	Odd - $\frac{3}{7}$
Odd - $\frac{2}{6}$	Even	Even
Odd - $\frac{1}{5}$	Even	Even

$\frac{5!}{P(9,5)} = \frac{5 \cdot 4\cdot 3 \cdot 2 \cdot 1}{9 \cdot 8\cdot 7 \cdot 6 \cdot 5} = \frac{1}{126}$

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 $9$ , we get:

$\frac{1}{126} \cdot 9 = \frac{9}{126} = \boxed{\textbf{(B) }\frac{1}{14}}$ -mgher

Solution 5

To get an odd sum we need $odd + odd + odd$ or we need $odd + even + even.$ Note that there are $9!$ ways to arrange the numbers. Also notice that there are $4$ even numbers and $5$ odd numbers from $1$ through $9$ .

The number of ways to arrange the even numbers is $4!$ and the number of ways to arrange the odd numbers is $5!$ . 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 $4$ even numbers we need to split two evens to one row and two evens to another row. Note that there are $\tbinom{3}{2}$ ways to choose the rows, and the $\tbinom{3}{2}$ 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 $9$ 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 $4!$ and $5!$ ways respectively.

Therefore our answer is simply $\frac{5! \cdot 4! \cdot 9}{9!} = \frac{1}{7\cdot 2} = \boxed{\frac{1}{14}}.$

Video Solution

For those who want a video solution:

1. https://www.youtube.com/watch?v=uJgS-q3-1JE

2. https://www.youtube.com/watch?v=3Sj7XPernCY

3. https://www.youtube.com/watch?v=APKGHtj-2rI

@@ Line 36: / Line 36: @@
 <cmath>\frac{1}{126} \cdot 9 = \frac{9}{126} = \boxed{\textbf{(B) }\frac{1}{14}}</cmath>
 -mgher
+==Solution 5==
+To get an odd sum we need <math>odd + odd + odd</math> or we need <math>odd + even + even.</math> Note that there are <math>9!</math> ways to arrange the numbers. Also notice that there are <math>4</math> even numbers and <math>5</math> odd numbers from <math>1</math> through <math>9</math>.
+The number of ways to arrange the even numbers is <math>4!</math> and the number of ways to arrange the odd numbers is <math>5!</math>. 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 <math>4</math> even numbers we need to split two evens to one row and two evens to another row. Note that there are <math>\tbinom{3}{2}</math> ways to choose the rows, and the <math>\tbinom{3}{2}</math> 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 <math>9</math> 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 <math>4!</math> and <math>5!</math> ways respectively.
+Therefore our answer is simply <cmath>\frac{5! \cdot 4! \cdot 9}{9!} = \frac{1}{7\cdot 2} = \boxed{\frac{1}{14}}.</cmath>
 ==Video Solution==

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

Page

Toolbox

Search