AMC10 2000 A

AMC10 2000 A · Q13

AMC10 2000 A · Q13. It mainly tests Logic puzzles.

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?

有一个三角形钉板，要放置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。有多少种方法可以放置这些钉，使得没有（水平）行或（垂直）列包含相同颜色的两个钉？

(A) 0 0

(B) 1 1

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

(E) 15! 15!

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): To avoid having two yellow pegs in the same row or column, there must be exactly one yellow peg in each row and in each column. Hence, starting at the top of the array, the peg in the first row must be yellow, the second peg of the second row must be yellow, the third peg of the third row must be yellow, etc. To avoid having two red pegs in some row, there must be a red peg in each of rows 2, 3, 4, and 5. The red pegs must be in the first position of the second row, the second position of the third row, etc. Continuation yields exactly one ordering that meets the requirements, as shown.

答案（B）：为避免在同一行或同一列出现两个黄色棋子，每一行和每一列必须恰好有一个黄色棋子。因此，从阵列顶部开始，第一行中的棋子必须是黄色，第二行的第二个棋子必须是黄色，第三行的第三个棋子必须是黄色，依此类推。为避免在某一行中出现两个红色棋子，第2、3、4、5行中每一行都必须有一个红色棋子。红色棋子必须位于第二行的第一个位置、第三行的第二个位置，等等。继续推下去可得到恰好一种满足要求的排列，如图所示。

Topics

LM.001 Logic puzzles