Circular Permutations

[unkempt rapids]

A round table meeting is held between eight persons A B C D E F G and H. In how many ways can they be seated so that A and H always sit diagonally opposite to each other?

Please tell me that the answer is 720.

[fast girder]

Yeah I think you are right

[jade panther]

Are you asking for any case where A is not directly next to H, or are you asking for diametrically opposite seating?
For the latter you get 8 general positions for A, H to be in, and since you can rearrange the other people B, C, D, E, F and G freely, resulting in 6! ways to seat everyone for each position of A. So your total answer would be
6!×8=6×5×4×3×2×8=720×8=5760

For the former option (i. e. A is not next to H), you get 5 seating options for H per position of A. Since there are 8 possible positions for A, H, you get 5×8=40 possible seatings for A and H alone.
Since you can rearrange the other people as shown above per position of A, H, you get a answer of
6!×40=720×40=28800

[fast girder]

Don’t forget that the order around the table don’t count

They can turn around and the set will remain the same

[jade panther]

In that case, 720 is correct for the first case, while the second case has a solution of 720×5=3600

[fast girder]

For the first yes it is

And for the second also you are right

Your reasoning is the exact correct one

[unkempt rapids]

I'm asking for diametrically opposite seating

[jade panther]

Then the answer is 720

[unkempt rapids]

Yes, I also believe that the answer is 720. But the solution in the book said 1440, lol. That's outright wrong.

[jade panther]

.solved

[tawny trench]

Solved

Post marked as solved by @jade panther.

Use .unsolved if this was a mistake.

[unkempt rapids]

Oh, is it compulsory to do?

Is this command available for me?

think it should be available only for me

[jade panther]

You just type .solved when a problem is done and .unsolved if it isn't solved after all.