Question

有多少个 $4 \times 4$ 的 $0$ - $1$ 数组，使得四行的和以某种顺序为 $1, 2, 3$ 和 $4$ 四列的和也以某种顺序为 $1, 2, 3$ 和 $4$ ？例如，数组满足条件。 $\begin{bmatrix} 1 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 1 & 1 & 1 & 1 \\ 0 & 1 & 0 & 0 \end{bmatrix}$

How many $4 \times 4$ arrays whose entries are $0$ s and $1$ s are there such that the row sums (the sum of the entries in each row) are $1, 2, 3,$ and $4,$ in some order, and the column sums (the sum of the entries in each column) are also $1, 2, 3,$ and $4,$ in some order? For example, the array $\begin{bmatrix} 1 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 1 & 1 & 1 & 1 \\ 0 & 1 & 0 & 0 \end{bmatrix}$ satisfies the condition.

$144$

$240$

$336$

$576$

$624$

答案：D

知识点：乘法原理分类讨论

难度评级：1840

解答：

行和为 $4$ 的行全是 $1$ ，列和为 $4$ 的列也全是 $1$ 。将行和 $1, 2, 3, 4$ 分配给四行有 $4!$ 种方式，选择哪一列的和为 $4$ 有 $4$ 种方式。

删除该列后，剩下的 $4 \times 3$ 数组的行和为 $0, 1, 2, 3$ ，列和必须为 $1, 2, 3$ 。全零行和全一行已确定；行和为 $1$ 和 $2$ 的两行有 $6$ 种排法，使列和按某种顺序成为 $1, 2, 3$ 。

因此总数为 $24 \cdot 4 \cdot 6 = 576$ 。

所以正确答案是 D。

The row with sum $4$ is all $1$ s and the column with sum $4$ is all $1$ s. There are $4!$ ways to assign the row sums $1, 2, 3, 4$ to the four rows, and $4$ choices for which column has sum $4.$

Delete that column. The remaining $4 \times 3$ array has row sums $0, 1, 2, 3$ and must have column sums $1, 2, 3.$ The all-zero and all-one rows are forced; the rows of reduced sum $1$ and $2$ can be placed in $6$ ways to produce column sums $1, 2, 3$ in some order.

The total is $24 \cdot 4 \cdot 6 = 576.$

Thus, the correct answer is D.

2022 AMC 12B 第 17 题

答案：D

难度评级：1840

解答：

其他年份的第 17 题