There are 18 chairs numbered from 1 through 18 around a circular table. How many ways can three people be seated, so that no two people are adjacent?
We can use a stars-and-bars technique. First, we cut the circle at a certain point, to get a row of chairs. Using stars-and-bars, there are C(16,3) = 560 ways to choose two chairs so that no two chairs are adjacent. Therefore, there are 560*6 = 3360 ways to seat the three people.
I tried this before an got in a mess. This time I am more confident.
But there is still plenty of room for error.
Seat each person and tie their chair to the chair on the right (which will be empty.)
Think of the 2 chairs and the one person as a single unit.
So now you have 3 fat units and 12 more chairs.
You have to slip the fat units between the other chairs.
15 units altogther and you need to choose the 3 that are fat.
so that is 15C3= 455
Now the three people can be in any order so that is 3!=6 ways
I tried both 3360 and 2730, and both are incorrect. Help!
What abuit if you put the first fat person anywhere. 15 places.
then there are 14 places to choose from and 2 have to be fat people
Slightly different method but still I get the same result.
(The logic is not very different)