2014 AMC 12A 第 3 题

先试着解答 2014 AMC 12A 第 3 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2014 AMC 12A 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

3.

Ralph 沿 Jane 街走,依次经过四栋房子,每栋房子颜色不同。他先经过橙色房子,再经过红色房子;并且先经过蓝色房子,再经过黄色房子。蓝色房子不与黄色房子相邻。彩色房子的排列顺序有多少种可能?
Walking down Jane Street, Ralph passed four houses in a row, each painted a different color. He passed the orange house before the red house, and he passed the blue house before the yellow house. The blue house was not next to the yellow house. How many orderings of the colored houses are possible?
22
33
44
55
66
答案:B
知识点:有限制的排列分类讨论
小提示:
按橙色房子先出现还是蓝色房子先出现分类。
Split into cases by whether the orange house or the blue house comes first
大提示:
因为蓝色和黄色不能相邻,所以它们之间至少隔一栋房子。
Because blue and yellow cannot be adjacent, at least one house must separate them
解答:
若橙色先出现,则蓝色和黄色不能相邻,迫使顺序为橙、蓝、红、黄。
若蓝色先出现,则黄色可以在第三或第四个位置(不能在第二个位置以避免相邻),得到蓝、橙、黄、红和蓝、橙、红、黄。
这就是全部 33 种有效顺序。
所以正确答案是 B。
If orange comes first, then blue and yellow cannot be adjacent, forcing the order orange, blue, red, yellow.
If blue comes first, yellow can be in the third or fourth position (never second, to avoid adjacency), giving blue, orange, yellow, red and blue, orange, red, yellow.
These are the only 33 valid orderings.
Thus, the correct answer is B.
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