Math Help - Seating arrangements

1. Seating arrangements

The question has four parts:
A. Suppose there are 8 seats in a row with 8 people to seat in the seats, one per seat, how many seating arrangements are possible?

B. Suppose 2 of the 8 are twins and are indistinguishable, thus any arrangement should be labeled the same regardless of which twin in is which seat, how many arrangements are possible?

C. Same as B but with triplets instead of twins

D. Same as B but with triplets and a different set of twins, still 8 total.

For A it should be 8! right? idk what to do for B, C, and D though, any help would be appreciated, TIA.

2. Originally Posted by jason0
Suppose there are 8 seats in a row with 8 people to seat in the seats, one per seat, how many seating arrangements are possible?
C. Same as B but with triplets instead of twins
If we rearrange the string "AAABCDEF" is that not the exact same as part C?
The number of possible arrangements is $\frac{8!}{3!}$.
Now you show us the rest.

3. Originally Posted by Plato
If we rearrange the string "AAABCDEF" is that not the exact same as part C?
The number of possible arrangements is $\frac{8!}{3!}$.
Now you show us the rest.
so would it be like
part a is ABCDEFGH so 8! = 40320 combinations
part b is AABCDEFG so 8!/2! = 20160 combinations
part c is AAABCDEF so 8!/3! = 6720 combinations
part d is AAABBCDE so 8!/(3!+2!) = 3360 combinations?

4. Originally Posted by jason0
part d is AAABBCDE so 8!/(3!+2!) = 3360 combinations?
That should be $\frac{8!}{(2!)(3!)}$

5. Originally Posted by Plato
That should be $\frac{8!}{(2!)(3!)}$
thank you