MHB Can You Help Me Count the Different Ways to Organize My List?

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The discussion focuses on organizing a list of songs liked by three girls under specific constraints. Each girl must like either two or three songs, ensuring that no song is liked by all three. Various cases are analyzed to determine the number of ways to select songs while adhering to these rules. The total number of different ways to organize the songs, considering all cases, is calculated to be 132. The conversation emphasizes the importance of careful case consideration to avoid contradictions in song preferences.
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First I tried to make an organized list but I kept on messing up.
Then I tried to subtract the cases that don't work but this also was not a good approach.
Could I get some help on setting up the cases I should be considering?
 

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veronica1999 said:
First I tried to make an organized list but I kept on messing up.
Then I tried to subtract the cases that don't work but this also was not a good approach.
Could I get some help on setting up the cases I should be considering?

Hi veronica1999, :)

Under the given constraints each girl should have 2 or 3 songs that she likes. If a girl likes only one of the songs, then the following condition cannot be satisfied.

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.

Similarly, if a girl likes all four of the songs, then the condition,

No song is liked by all three.

cannot be satisfied.

Let me name the girls as A, B and C. Now consider each case,

[TABLE="class: grid, width: 550, align: center"]
[TR]
[TD="align: center"][/TD]
[TD="align: center"]No. of songs liked by A[/TD]
[TD="align: center"]No. of songs liked by B[/TD]
[TD="align: center"]No. of songs liked by C[/TD]
[/TR]
[TR]
[TD="align: center"]1)[/TD]
[TD="align: center"]3[/TD]
[TD="align: center"]3[/TD]
[TD="align: center"]2[/TD]
[/TR]
[TR]
[TD="align: center"]2)[/TD]
[TD="align: center"]3[/TD]
[TD="align: center"]2[/TD]
[TD="align: center"]3[/TD]
[/TR]
[TR]
[TD="align: center"]3)[/TD]
[TD="align: center"]2[/TD]
[TD="align: center"]3[/TD]
[TD="align: center"]3[/TD]
[/TR]
[TR]
[TD="align: center"]4)[/TD]
[TD="align: center"]3[/TD]
[TD="align: center"]2[/TD]
[TD="align: center"]2[/TD]
[/TR]
[TR]
[TD="align: center"]5)[/TD]
[TD="align: center"]2[/TD]
[TD="align: center"]3[/TD]
[TD="align: center"]2[/TD]
[/TR]
[TR]
[TD="align: center"]6)[/TD]
[TD="align: center"]2[/TD]
[TD="align: center"]2[/TD]
[TD="align: center"]3[/TD]
[/TR]
[TR]
[TD="align: center"]7)[/TD]
[TD="align: center"]2[/TD]
[TD="align: center"]2[/TD]
[TD="align: center"]2[/TD]
[/TR]
[/TABLE]

If we consider the first case,

'A' likes 3 songs out of 4. There are, \({}^4C_{3}\) ways to choose these three songs. 'B' also likes 3 songs out of 4. Suppose 'B' likes the same three songs that 'A' likes. Then 'C' should like a song that both 'A' and 'B' like. This cannot happen as it is given that,

No song is liked by all three.

Therefore, 'B' should like only two songs that 'A' like, and the other one is the one that 'A' dislikes. The number of ways to choose the two songs(out of the 3 that A likes) is given by, \({}^3C_{2}\). Now if you think carefully you will see that only a pair of songs are left for 'C' to like, without violating the given criteria.

Therefore the total number of possibilities for the first case \(={}^4C_{3}\times{}^3C_{2}\)

If the above explanation is hard to visualize the diagram that I have attached may help. The fours songs are denoted by 1,2,3 and 4.

Likewise I have considered each case separately. These are given in the following table.[TABLE="class: grid, width: 550, align: center"]
[TR]
[TD="align: center"][/TD]
[TD="align: center"]No. of songs liked by A[/TD]
[TD="align: center"]No. of songs liked by B[/TD]
[TD="align: center"]No. of songs liked by C[/TD]
[TD="align: center"]No. of Ways to choose the songs[/TD]
[/TR]
[TR]
[TD="align: center"]1)[/TD]
[TD="align: center"]3[/TD]
[TD="align: center"]3[/TD]
[TD="align: center"]2[/TD]
[TD="align: center"]\({}^4C_{3}\times{}^3C_{2}\)[/TD]
[/TR]
[TR]
[TD="align: center"]2)[/TD]
[TD="align: center"]3[/TD]
[TD="align: center"]2[/TD]
[TD="align: center"]3[/TD]
[TD="align: center"]\({}^4C_{3}\times{}^3C_{1}\)[/TD]
[/TR]
[TR]
[TD="align: center"]3)[/TD]
[TD="align: center"]2[/TD]
[TD="align: center"]3[/TD]
[TD="align: center"]3[/TD]
[TD="align: center"]\({}^4C_{2}\times{}^2C_{1}\)[/TD]
[/TR]
[TR]
[TD="align: center"]4)[/TD]
[TD="align: center"]3[/TD]
[TD="align: center"]2[/TD]
[TD="align: center"]2[/TD]
[TD="align: center"]\({}^4C_{3}\times{}^3C_{1}\times{}^2C_{1}\)[/TD]
[/TR]
[TR]
[TD="align: center"]5)[/TD]
[TD="align: center"]2[/TD]
[TD="align: center"]3[/TD]
[TD="align: center"]2[/TD]
[TD="align: center"]\({}^4C_{2}\times{}^2C_{1}\times{}^2C_{1}\)[/TD]
[/TR]
[TR]
[TD="align: center"]6)[/TD]
[TD="align: center"]2[/TD]
[TD="align: center"]2[/TD]
[TD="align: center"]3[/TD]
[TD="align: center"]\({}^4C_{2}\times{}^2C_{1}\times{}^2C_{1}\)[/TD]
[/TR]
[TR]
[TD="align: center"]7)[/TD]
[TD="align: center"]2[/TD]
[TD="align: center"]2[/TD]
[TD="align: center"]2[/TD]
[TD="align: center"]\({}^4C_{2}\times{}^2C_{1}\times{}^2C_{1}\)[/TD]
[/TR]
[/TABLE]

Therefore the total number of different ways \(=\left({}^4C_{3}\times{}^3C_{2}\right)+\left({}^4C_{3}\times{}^3C_{1}\right)+\left({}^4C_{2}\times{}^2C_{1}\right)+\left({}^4C_{3}\times{}^3C_{1} \times{}^2C_{1}\right)+3\left({}^4C_{2}\times{}^2C_{1}\times{}^2C_{1}\right)=132\)

Kind Regards,
Sudharaka.
25035mh.png

 
Last edited:
Thank you!
You are really awesome.:D
 
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