Seating 2 people in a row **There are 27 seats in a row. What's the probability of seating Jim and Dan in a way that there's no more than one vacant space between them?** So far I've reasoned that you can seat 2 people in 27 seats in $\binom{27}{2}$ ways, so that would be the number of possible events. The acceptable events for me are the ones where Jim and Dan are sitting next to each other, and the ones where Jim and Dan are sitting one space away from each other. If I put Jim first at one of the ends, he can take one of 27 seats then Dan can take only one seat. If I put Jim first non-end spaces, Dan has two options for the seats. And this is where I'm kinda stuck... I'm not sure how to proceed. I have no clue how to approach the one-seat-apart problem at all. Hints?

It's more simple than you are making it.

You can choose any two adjacent seats (no gap). There are $26$ such pairs.

You can choose any three adjacent seats (one gap). There are $25$ such triples.

So there are $51$ total positionings, but they can be flipped, giving a true total of $102$.