AMC10 2012 B

AMC10 2012 B · Q24

AMC10 2012 B · Q24. It mainly tests Basic counting (rules of product/sum), Inclusion–exclusion (basic).

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?

Amy、Beth 和 Jo 听了四首不同的歌曲，并讨论她们喜欢哪些。没有一首歌被三人全部喜欢。而且，对于女孩的三对中的每一对，至少有一首歌被这两人喜欢但第三人不喜欢。有多少种不同的情况可能？

(A) 108 108

(B) 132 132

(D) 846 846

(E) 1105 1105

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): There are two cases to consider. Case 1 Each song is liked by two of the girls. Then one of the three pairs of girls likes one of the six possible pairs of songs, one of the remaining pairs of girls likes one of the remaining two songs, and the last pair of girls likes the last song. This case can occur in $3\cdot 6\cdot 2=36$ ways. Case 2 Three songs are each liked by a different pair of girls, and the fourth song is liked by at most one girl. There are $4!=24$ ways to assign the songs to these four categories, and the last song can be liked by Amy, Beth, Jo, or no one. This case can occur in $24\cdot 4=96$ ways. The total number of possibilities is $96+36=132$.

答案（B）：需要考虑两种情况。情况 1 每首歌都被两个女孩喜欢。则三个女孩配对中的某一对喜欢六种可能的“歌曲对”中的一种，剩下的女孩配对中有一对喜欢剩下的两首歌中的一首，最后一对女孩喜欢最后一首歌。此情况共有 $3\cdot 6\cdot 2=36$ 种。情况 2 有三首歌分别被不同的一对女孩喜欢，第四首歌最多只被一个女孩喜欢。将四首歌分配到这四类的方式有 $4!=24$ 种，而最后那首歌可以被 Amy、Beth、Jo 喜欢，或无人喜欢。此情况共有 $24\cdot 4=96$ 种。总的可能数为 $96+36=132$。

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