# Seating arrangements

• Feb 19th 2009, 01:27 PM
jason0
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.
• Feb 19th 2009, 01:41 PM
Plato
Quote:

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 $\displaystyle \frac{8!}{3!}$.
Now you show us the rest.
• Feb 19th 2009, 01:48 PM
jason0
Quote:

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 $\displaystyle \frac{8!}{3!}$.
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?
• Feb 19th 2009, 02:03 PM
Plato
Quote:

Originally Posted by jason0
part d is AAABBCDE so 8!/(3!+2!) = 3360 combinations?

That should be $\displaystyle \frac{8!}{(2!)(3!)}$
• Feb 19th 2009, 02:03 PM
jason0
Quote:

Originally Posted by Plato
That should be $\displaystyle \frac{8!}{(2!)(3!)}$

thank you