AMC12 2019 A

AMC12 2019 A · Q16

AMC12 2019 A · Q16. It mainly tests Probability (basic), Casework.

The numbers 1, 2, ..., 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?

数字1, 2, ..., 9被随机放置到一个$3 \times 3$网格的9个方格中。每个方格得到一个数字，且每个数字只使用一次。每个行和每个列的数字之和为奇数的概率是多少？

(A) \frac{1}{21} \frac{1}{21}

(B) \frac{1}{14} \frac{1}{14}

(D) \frac{2}{21} \frac{2}{21}

(E) \frac{1}{7} \frac{1}{7}

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): The sum of three integers is odd exactly when either all of the integers are odd, or one is odd and two are even. Five of the numbers 1, 2, ..., 9 are odd, so at least one row must contain two or more odd numbers. Thus one row must contain three odd numbers and no even numbers, and the other two rows must contain one odd number and two even numbers. The same is true of the three columns. There are $3\times 3=9$ ways to choose which row and which column contain all odd numbers, and then the remaining four squares must have even numbers. There are $\binom{9}{4}=126$ ways in total to choose which squares have odd numbers and which have even numbers, so the desired probability is $\frac{9}{126}=\frac{1}{14}$.

答案（B）：三个整数之和为奇数，当且仅当要么三个整数全为奇数，要么一个为奇数、两个为偶数。数字 $1,2,\ldots,9$ 中有 5 个是奇数，因此至少有一行必须包含不少于两个奇数。于是必有一行包含三个奇数且不含偶数，而另外两行各包含一个奇数和两个偶数。三列同理。选择哪一行和哪一列全为奇数共有 $3\times 3=9$ 种方式，此时剩下的四个格子必须为偶数。总共有 $\binom{9}{4}=126$ 种方式选择哪些格子放奇数、哪些放偶数，所以所求概率为 $\frac{9}{126}=\frac{1}{14}$。

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