In a rectangular table with two heads of one seat each and three seats on each side, in how many ways can we sit a family with 2 parents and 6 children, with the parents on the opposing heads of the table?
My answer was $2! \times 6! = 1440$, given that we should have $2!$ ways of seating the parents on the opposing heads and $6!$ ways of seating the children. However, the book answer is $1140$. I do not know where this comes from. I thought they were considering the heads as similar and thus when we exchanged the parents you could have repeated configurations. Yet, in that case, each of these $1440$ configurations would be counting twice the same configuration, therefore we should have $720 = \frac{1440}{2}$ instead. However, the next question in the book is how many ways could I seat the family so that two specific children were facing each other; in this question, their answer is $5760$, which could be reached by doing $4 \times 2! \times 6! = 5760$. This answer clearly indicated that each head is unique and hence so is every other seat! Please help.
Thank you very much. All the best.
1140 is definitely a typo. The explanation of the factors in $4×2!×6!$ is that there are 4 ways to choose the opposing seats where the two specific children are, $2!$ ways to permute those children and $6!$ ways to permute the rest of the family.