manlog wrote:

I solved it in a bit simpler form:

There are 10 spots for 10 people. How can we break the requirement? What's the probability of arranging these two people in a way that they sit next to each other?

Probability of choosing a person is 1, then what's the probability of choosing a person sitting next to him = 1/9 * 2 (since there are two seats right next to him). Then the probability of arranging these two people in a way, that they don't seat to each other is 1 - 2/9 = 7/9

My approach is very similar -

1. Let's say Ann sits at a place.

2. Probability that Bill sits next to Ann in circular arrangement is \(\frac{[Left + Right]}{[Remaining 9 places]}\) = \(\frac{2}{9}\)

3. Probability of both not sitting together is \(1 - \frac{2}{9} = 7/9\)