Ok so here's the problem...

Suppose there are N people, and N chairs to seat them, all in a row. There are, however, 2 people (who we will call bob and row) who will only sit next to each other. That means that if bob sits in the 11th seat, then row must sit next to him, in either the 10th or 12th. this also means that if one of them (bob or rowdy, but we'll just say bob for example) is sitting in either of the last seats, then rowdy must sit in either n-1 if bob is sitting in the n'th seat (since n+1 doesn't exist) or in 2 if bob is sitting in 1 (since 0 doesn't exist).

so, how many possible combinations of N people, if 2 wish to sit next to each other.

I couldn't figure out a logical answer, but after allot of enumeartion (which i know is not the correct way to do things, but it can sometimes steer you in the right way) i came up with $\displaystyle n!/(n-2)!$

however, when i try to prove this with induction (the only way we know how to formally prove currently) i get stuck when trying n+1 with :

$\displaystyle (n!/(n-2)!) * (n+1)/(n-1)$

and i can't figure out how to logicaly deduce that $\displaystyle (n+1)/(n-1)$ will give us the next set of iterations.

anyway, the main question is the first one about number of combinations, the second is just a curiosity- since i'm sure we weren't supposed to prove this using induction. Thanks allot! I really appreciate your help, and if you could, please explain! it's probably very simply, but I just can't put my finger on it.

Thanks!