AMC10 2019 A

AMC10 2019 A · Q20

AMC10 2019 A · Q20. It mainly tests Probability (basic), Parity (odd/even).

The numbers 1, 2, ..., 9 are randomly placed into the 9 squares of a 3 × 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 随机放置到 3 × 3 网格的 9 个方格中。每个方格放一个数字，每个数字使用一次。每行和每列数字之和均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 $1,2,\ldots,9$ 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两个或更多奇数。于是必有一行包含三个奇数且不含偶数，而另外两行必须各包含一个奇数和两个偶数。三列同理。有$3\times 3=9$种方法选择哪一行和哪一列全为奇数；此时剩下的四个格子必须为偶数。总共有$\binom{9}{4}=126$种方法选择哪些格子放奇数、哪些格子放偶数，所以所求概率为$\frac{9}{126}=\frac{1}{14}$。

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