Recurrence in sight?

Firstly, assume that there is someone in the middle of the row (seat $N$ ), who chooses to sit in seat $N-1$ . This forces the person in seat $N-1$ to change seats. If they sit in seat $N-2$ , then surely this process repeats, with each person in seat $k$ forced to sit in seat $k-1$ . This would therefore create a problem as the person assigned seat $1$ would then have nowhere to sit.

Therefore, for every person assigned to seat $a$ that chooses to sit in seat $b$ , the person assigned seat $b$ must sit in seat $a$ .

Now we have to consider how many pairs of people choose to swap seats with somebody next to them. The number of pairs of seat-swappers is at least 0, and at most 5. If there were 0 pairs, there would be $\binom{10}{0} = 1$ possibility (everyone sitting in their assigned seat). For $m$ pairs of swappers, there would be $\binom{10-m}{m}$ .

Therefore, $\binom{10}{0} + \binom{9}{1} + \binom{8}{2} + \binom{7}{3} + \binom{6}{4} + \binom{5}{5} = 1 + 9 + 28 + 35 + 15 + 1$ $\binom{10}{0} + \binom{9}{1} + \binom{8}{2} + \binom{7}{3} + \binom{6}{4} + \binom{5}{5} = 89$

Let $F_n$ be the number of ways in which $n$ students can sit in $n$ seats so that each student sits either in the chair specified on his/her ticket or in one right next to it.

Case 1: When the leftmost student sits on the leftmost seat (his seat on his ticket), observe that the problem changes to $n-1$ students and $n-1$ seats (the other conditions remain the same). Thus there are $F_{n-1}$ seating arrangements for his friends.

Case 2: When the leftmost student sits on sits on the seat right of that assigned to him on the ticket, observe that the person right to him has to sit (no other possibility) on the left seat (the seat of the leftmost person). Now the problem changes to $n-2$ students and $n-2$ seats (the other conditions remain the same). Thus there are $F_{n-2}$ seating arrangements for his friends.

Thus we come to the conclusion $F_n=F_{n-1}+F_{n-2}$ and this looks familiar , but in this case you will find that $F_3=3$ and $F_4=5$ (you can work it out for smaller values of $n$ ). Using the recurrence and the two known terms, you can go up till $10$ and you will find that $F_{10}= \boxed {89}$