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 peopl

Total number of ways people can be seated:

18 choose 3

\(\frac{18!}{3!15!}=816\)

Since no two people can be adjacent we have two cases

Three people adjacent

18 ways (18 choose 1)

Two people adjacent (18 * 14)

So 816 - 18 - (18*4)

546

Is this right?

Let me read over your work, and check. Thanks for your effort, I appreciate it! Thanks!

There are 18*17*16 ways to choose the three seats, if people can be adjacent.  We must subtract the cases where there are two people adjacent, which is 18*13*6.  So the number of ways is 18*17*16 - 18*13*6 = 3492.

Well, isn't there 18*17*16 ways to pick? 18 choices for the first person, etc... thats 4896. Could you explain why there are 18*14 cases of 2 people adjacent? Thanks again!

Alright, the answer is 3276. Without any restrictions, there a total of  ways that the three people can be seated. We must then subtract the cases where at least two people are adjacent. There are two possible cases.

Case 1. All three people are sitting next to each other, in one block.

There are 18 ways to choose three consecutive seats. There are then 3! = 6ways to place the three people, which gives us 18 * 6 =108 possible seatings.

Case 2. There are two people sitting next to each other, and the third person is not next to either of them.

There are 18 ways to choose two consecutive seats. There are then 14 ways to choose the third seat. After the three seats have been chosen, there are  ways to place the three people, which gives us 18*14*6=1512 possible seatings.

Therefore, there are 4892-108-1512 = 3276 seatings where no two people are adjacent.

Thank you all for the effort, it's greatly appreciated!