Question

有 $5$ 个黄色钉、 $4$ 个红色钉、 $3$ 个绿色钉、 $2$ 个蓝色钉和 $1$ 个橙色钉，要放在一个三角形钉板上。要求任意水平行或竖直列中不能有两个同色钉。共有多少种放法？

There are $5$ yellow pegs, $4$ red pegs, $3$ green pegs, $2$ blue pegs, and $1$ orange peg to be placed on a triangular peg board. In how many ways can the pegs be placed so that no (horizontal) row or (vertical) column contains two pegs of the same color?

$0$

$1$

$5! \cdot 4! \cdot 3! \cdot 2! \cdot 1!$

$15!/(5! \cdot 4! \cdot 3! \cdot 2! \cdot 1!)$

$15!$

答案：B

知识点：有限制的排列逻辑推理

难度评级：1370

解答：

钉板有五行和五列。为了避免同一行或同一列中出现两个黄色钉，每行和每列都必须恰有一个黄色钉，因此五个黄色钉的位置被确定在长对角线上。

接着，四个红色钉必须分别放在第 $2$ 行至第 $5$ 行中，剩余的位置也迫使它们排在一条对角线上。继续放置绿色、蓝色和橙色钉时，每种颜色的位置也都被唯一确定。

因此恰好有一种符合条件的放法。

所以正确答案是 B。

The board has five rows and five columns. To avoid two yellow pegs in a row or column, there must be exactly one yellow peg in each row, forcing the yellow pegs onto the long diagonal.

The four red pegs must then each go in rows $2$ through $5,$ and the only positions left force them into a single diagonal as well. Continuing with green, blue, and orange, every color is forced into a unique position.

Hence there is exactly one valid arrangement.

Thus, the correct answer is B.

2000 AMC 10 第 13 题

答案：B

难度评级：1370

解答：

其他年份的第 13 题