Question

Amy、Beth 和 Jo 听了四首不同的歌，并讨论她们喜欢哪些歌。没有一首歌被三个人都喜欢。此外，对三个人中的每一对，都至少有一首歌被这两个人喜欢，而第三个人不喜欢。有多少种不同的可能情况？

Amy, Beth, and Jo listen to four different songs and discuss which ones they like. No song is liked by all three. Furthermore, for each of the three pairs of the girls, there is at least one song liked by those two girls but disliked by the third. In how many different ways is this possible?

$108$

$132$

$671$

$846$

$1105$

答案：B

知识点：基本计数分类讨论乘法原理

难度评级：2100

解答：

分两种情况：某一对女孩共同喜欢 $2$ 首歌，或每一对恰好有 $1$ 首共同喜欢的歌；若共同配对更多，就至少需要 $5$ 首歌。

第一种情况：三对女孩各自恰好有一首只被这对喜欢的歌。

给第一对选歌有 $4$ 种，第二对有 $3$ 种，第三对有 $2$ 种；剩下的歌可以被 $3$ 人中的恰好一人喜欢或无人喜欢。因此这一类有 $4\cdot 3\cdot 2\cdot (3+1)=96.$

接着考虑第 $2$ 种情况。

选共同喜欢两首歌的那一对有 $3$ 种；从四首歌中选出这对喜欢的 $2$ 首有 $\binom 42 = 6$ 种；再把剩下两首分配给另外两对，有 $2! =2$ 种。因此这一类有 $3\cdot 6\cdot 2=36$

总数为 $96+36 = 132$ 。

所以正确答案是 B。

There are two cases: Each pair has exactly one liked song in common, or some pair has $2$ liked songs in common. This is because each pair must have at least $1$ liked song in common, and any more pairs than in the cases would result in $5$ songs.

Case 1: Each pair has exactly one liked song in common

There are $4$ ways to choose the song that one pair likes, $3$ ways to choose the song that the second pair likes, and $2$ ways to choose the song the third pair likes if we choose some order for them. Then, for the last song, one of them could like it which has $3$ cases or none of them likes it which is another case. Thus, the number of solutions in this case is $4\cdot 3\cdot 2\cdot (3+1)=96.$

Case 2: Some pair has $2$ liked songs in common

There are $3$ ways to choose the pair that has $2$ liked songs in common. Then, there are $\binom 42 = 6$ ways to choose which two songs they like. Finally, there are $2! =2$ ways to assign the two remaining songs to the other two pairs. Thus, the number of solutions in this case is $3\cdot 6\cdot 2=36$

The total amount is then $96+36 = 132.$

Thus, the correct answer is B .

2012 AMC 10B 第 24 题

答案：B

难度评级：2100

解答：

其他年份的第 24 题