用四种不同颜色给图中ABCDEF六个点涂色,每点涂一色,每条线段两端点颜色不同,共几种涂法?

搞不懂 麻烦给详细点 这题好像很难
2025-04-02 00:21:52
推荐回答(1个)
回答1:

解法一: 按选用颜色种数进行分类. 【解析】分三类:(1)B、D、E、F用四种颜色,则有A必与F颜色相同、C与E颜色相同,故4A4×1×1=24种方法. (2)B、D、E、F用三种颜色,则有:B、E同色或D、F同色必有其一,若B、E同色,则A有异于B和D的两种颜色,C只有一种,D、F同色同理,1×2×2×4A3;B与D同色,则A、C都有异于B、E两种选择, 2×2×4A3,故96+96=192种. (3)B、D、E、F用二种颜色,只能B、E同色,D、F同色,A、C有异于B、D两种颜色,则有4A2×2×2=48,所以共有不同的涂色方法有24+192+48=264种.
解法二:利用“捆绑法”, 分步着色. 【解析】第一类:用三种颜色涂色,A、D、E颜色各不相同,若B与E同色,必有C与A、F与D同色,可将C与A看作一个整体,F与D看作一个整体;若B、D同色同理,故234×A种. 第二类:四种颜色(都用)涂六个点,必有4个点的位置颜色不同,即这六个点中必有两组点同色,看作一个整体,而这两组必为:AF、AC、BE、BD、CE、DF中的两组,如下:(AF、BE),(AF、BD),(AF、CE),(AC、BE),(AC、BD),(AC、DF),(BE、DF),(BD、CE),(CE、DF)共9种,944×A,共有不同的涂色方法有234×A+944×A=264种

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