# A question invoving combinatorics and many limitations [closed]

Three kids, Andrew, Basil and George, were listening to four different songs. None of the four songs was appreciated, by all three kids. For each of the three possible pairs of kids (Andrew and Basil or Andrew and George or Basil and George), there was at least one song, which was appreciated by both of kids. With how many different ways, can this occur? I was attempting this problem today, with no success. Can you guys please help me?

Thanking you in advance

Kevin

## closed as off-topic by Macavity, Mike Earnest, Leucippus, Shailesh, CesareoApr 18 at 7:30

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Macavity, Mike Earnest, Leucippus, Shailesh, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.

Case 1: One song is unappreciated.

Number of ways:

Choose one song to be unappreciated: $$\dbinom{4}{1}$$

From those remaining, one song will be appreciated by the first pair, one song will be appreciated by the second pair, and one by the third pair. There are $$3!$$ ways to permute them. So, this means there are $$4!$$ ways for one song to be unappreciated.

Case 2: One song is appreciated by exactly one child.

Number of ways:

Choose the one song, choose which child appreciates it: $$\dbinom{4}{1}\dbinom{3}{1} = 4\cdot 3$$

Next, there are $$3!$$ ways to permute the remaining three songs among the three children. So, there are $$3\cdot 4!$$ ways total.

Case 3: All four songs are appreciated by exactly 2 of the kids.

Choose two songs, choose one pair to like both of them: $$\dbinom{4}{2}\dbinom{3}{1} = 18$$

Next, permute the two remaining songs among the two remaining pairs, so there are a total of $$6\cdot 3!$$ ways of having all four songs be appreciated by exactly 2 of the kids each.

All told, there are $$4\cdot 4!+6\cdot 3! = 132$$ ways in which the outcome you are looking for can occur.