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Arrangments at a table

mark

Suppose that I sit down at the head of a table with all 10 of my discrete math students at the other 10 chairs.

How many different ways are there for this arrangement to happen?
Suppose all you care about is who sits next to who, so that rotational direction does not matter. Now, how many different arrangements are there?

ksimmon1

For the first part would the answer just be

(n-1)! = (11-1)! = 10! = 3628800

I am struggling with the second part of the problem.

mark

@ksimmon1 Here is my hint on part 2:

ksimmon1

Would we divide the total number we got from the #1 by 2 to account for the tables where everyone is still sitting next to the same people just in reverse order?

\frac{(n-1)!}{2} = \frac{(11-1)!}{2} = \frac{10!}{2} = \frac{3628800}{2} = 1814400