At a dinner party 5 people are to be seated arround

[Luci]

How would you do this one?

At a dinner party 5 people are to be seated arround a circular table. Tow seating arrangemets are considered different only when the positions of the people are different relative to each other. What is the total number of different possible seating arrangements for the group?

5,10,24,32,120

Answer is 24

[Guest]

Consider these as 5 seats around a round table.

5 people in a row will sit in 5*4*3*2*1 ways = 120 ways.

In circle, abcde = bcdea = cdeab = deabc = eabcd. So each different arrangement (by people's respective position) is actually reprsented 5 times in this 120. So divide by 5 and you get 24.

[Harish Dorai]

To add on to the point made by Guest - We can generalize this to any number of people. For any number N, the circular permutation formula is (N-1)!.

In the example, it is 5 people, so the answer is (5-1)! = 4! = 24.

[anadi]

and that was me as guest, forgot to put the id.

[jasonthomasyee]

Is this question solvable using the slot method?

[RonPurewal]

Is this question solvable using the slot method?

Yes"”"”as long as you either (a) understand how to create slots that won't lead to redundant outcomes, or (b) make five slots, but then get rid of the redundancy.

a/
Say the people are A, B, C, D, and E.
Since the table is round, you can take any configuration of guests and simply rotate the table so that guest "A" is seated at a certain position. (If "A" is already seated there, then there's no need to rotate.) So, for instance, BACDE, CDEBA, DEBAC, and EBACD are all the same outcome as ACDEB.
With this realization made, you can just put "A" in the designated position, and then make 4 slots that can be freely filled with the other four individuals: 4 x 3 x 2 x 1 = 24.

b/ see below

[RonPurewal]

b/
This is mostly the same deal that was already described earlier in the thread.
If you make 5 slots, then you'll get every configuration of guests exactly 5 times (e.g., the aforementioned BACDE, CDEBA, DEBAC, EBACD, and ACDEB, all of which represent the same seating arrangement).
So, your 5 slots are filled with 5 x 4 x 3 x 2 x 1, but then you need to divide by 5 so that every configuration is represented only once, rather than five times.
The 5's cancel, leaving the same answer as before.