8 people are seated in a row. What is the probability 2 people are at least 2 seats apart?

Question

My attempt: there are 5 ways they can have 2 seats between them, 4 ways they can have 3 seats between them, and so on, so that I got 5+4+3+2+1=15. multiply that by 2 because there are two of them, and multiply that by 6! for the remaining 6 people, so I got $$\frac{(5+4+3+2+1)\cdot 2\cdot 6!}{8!}=0.536$$ Is this right?

Answer 1

Yes, that's right. It would have been slightly easier to calculate the probability of the complement: There are $6$ ways to have $1$ seat between them and $7$ ways to have no seats between them, leading to

$$ 1-\frac{(6+7)\cdot2\cdot6!}{8!}\approx0.536\;. $$