Combinatorics Problem: Seating a Family of 8 with Twins and Siblings

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The problem involves seating the Jones family of 8, consisting of 5 boys and 3 girls, where 2 girls are twins who must sit together, while the third sister cannot sit next to them. The initial approach correctly identifies the arrangement of the twins as 2! times 7!, but fails to accurately account for the restriction on the third sister. To solve this, it is suggested to treat the twins as a single unit, allowing for a simpler arrangement with the boys and the third sister. The correct calculation involves considering the twins as one entity and carefully arranging the remaining family members to ensure the third sister is not adjacent to them.
Mr Davis 97
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Homework Statement


The Jones family has 5 boys and 3 girls, and 2 of the girls are twins. In how many ways can they be seated in a row of 8 chairs if the twins insist on sitting together, and their other sister refuses to sit next to either of her sisters?

Homework Equations

The Attempt at a Solution


I thought that I could use a "count the complement" technique. First, we would count the the number of ways to just have the two twins paired together. This would be ##2! \cdot 7!## ways. However, this over counts because it includes the pairs where the other sister is adjacent. Thus, we subtract from this ##3! \cdot 6!##, which is the number of arrangements where the other sister is adjacent to the other sisters. This gives 5760. However, this is not the right answer. What am I doing wrong?
 
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Mr Davis 97 said:

Homework Statement


The Jones family has 5 boys and 3 girls, and 2 of the girls are twins. In how many ways can they be seated in a row of 8 chairs if the twins insist on sitting together, and their other sister refuses to sit next to either of her sisters?

Homework Equations

The Attempt at a Solution


I thought that I could use a "count the complement" technique. First, we would count the the number of ways to just have the two twins paired together. This would be 2!⋅7!2!⋅7!2! \cdot 7! ways. However, this over counts because it includes the pairs where the other sister is adjacent. Thus, we subtract from this 3!⋅6!3!⋅6!3! \cdot 6!, which is the number of arrangements where the other sister is adjacent to the other sisters. This gives 5760. However, this is not the right answer. What am I doing wrong?

You are right about the number of ways to just have the two twins paired together (##2!\times 7!##). But then, to account for the third sister not being adjacent, you have to think more carefully. As a hint, I recommend to treat the three sisters together. Now, how many ways are there to arrange this with the boys? How many about the three sisters together?
 
Mr Davis 97 said:
we subtract from this 3!⋅6!,
Please explain your reasoning for that number. Remember, you have already combined the twins into one entity.
 
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