有16名学生参加一次竞赛。全是选择题,每题有四个选项。考完后发现任何两名学生的答案至多有一道题相同
推荐回答(3个)
题目说至多一题回答相同
但有16个学生 1道题目有4个选择。
那么 可以得出一点 一道题目一个选项 有且只有4个人选择。理由很简单。
因为至多只有一题答案相同 如果超出4人(别说少于4人,少的那个人不得选其他的。。)那么这一选项便是4+N 那么下一题其中一人选择A 那么剩下3个选择则有至少4个人选择,必定有两个人相同 那么这两人必定是两题答案相同 与题目相违背。所以
从题目中就可以得出这点结论 一道选项有且只有四人选择。
然后就好办了。
假设第一题 选择A的四人为 X1选B的四人为X2……
以A1为标准 他选择下一题的选项为X1Y1 那X1在第二题就分别为X1Y1 X1Y2 X1Y3 X1Y4
以下类推
(其实那Y可以取消掉的知道后面数字是什么意思就行了。)
然后根据题目 要求 除了最后一位 比如X1Y2 和X3Y2 X1Y3Z2和X3Y2Z2 前面的编号没有重复。
现在就好做了 只是简单的1234排列 无重复就行了。随便你怎么排 都不可能出现6位的数字 那么明显 至多只有5题
假设第一题 选择A的四人为 X1选B的四人为X2……
以A1为标准 他选择下一题的选项为X1Y1 那X1在第二题就分别为X1Y1 X1Y2 X1Y3 X1Y4
以下类推
(其实那Y可以取消掉的知道后面数字是什么意思就行了。)
然后根据题目 要求 除了最后一位 比如X1Y2 和X3Y2 X1Y3Z2和X3Y2Z2 前面的编号没有重复。
现在就好做了 只是简单的1234排列 无重复就行了。随便你怎么排 都不可能出现6位的数字 那么明显 至多只有5题
5道题的话如果16个人都选择A呢?那不是任意两个人都有5个相同选择?
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