I was working on two permutation questions, and wasn't sure if I arrived at the correct answer.
1.) In how many ways can a family of four (mother, father, and two children) be seated at a round table, with eight other people, so that the parents are seated next to each other with one child beside each parent?(Two seatings are considered the same if one can be rotated to look like the other).
The first thing I did was figure out the possible ways the family could sit:
Case 1: child #2 + mother + father + child #1
Case 2: child #1 + mother + father + child #2
Case 3: child #2 + father + mother + child #1
Case 4: child #1 + father + mother + child #2
Then I treated each case as one person and I calculated the # of ways this "one person" could be seated with 8 other people. Because it is a round table I did $(n-1)!= (9-1)! = 8!$. Since there are four cases I concluded that the total possible ways to seat the family, given the restrictions is $8! \cdot 4$. Not sure if this is right, but I feel that two of the cases are repeats because they are seated in a round table?
2.) Six people attend the theatre together and sit in a row with exactly six seats. There is an aisle on each end of the row. Suppose one of the six is a doctor who must sit on the aisle in case she is paged. How many ways can the people be seated together in the row with the doctor in an aisle seat?
For this question, I visualized the scenario as such: |_ _ _ _ _ _| So the doctor can either sit at the extreme left, in which case the total ways to be seated would be $5!$. Or she could be seated at the extreme right, in which case the total ways to be seated will also be $5!$. I added the two together $2 \cdot 5!$, to get a total of $240$ permutations, given the restrictions. Is that correct? Thanks!