Chew on this Gum...

Generalize this so $g_n$ is the number of ways this happens if there are n people.

Then, choose the nth person. If he sticks his gum under his own chair, then the case simplifies to that if there are n-1 people.

If he puts it underneath an adjacent chair, then the two people swap and it continues with n-2 people.

Then, we get $g_n=g_{n-1}+g_{n-2}$ .

We find $g_1=1$ and $g_2=2$ .

Finally, using our recursion, we find $g_{10}=\boxed{89}$ .

Perhaps the most important thing to realize here is that if person $A$ sticks his/her gum under the chair of $B$ then $B$ MUST stick his/her gum under the chair of $A$ if there is to be only one gum under every chair otherwise first/last person would have no gum under his/her chair. Thus a given person sticks his/her gum below his/her own chair or two people simply "exchange" their gums forming a pair. There can be minimum $0$ such pairs or maximum $5$ pairs. Consider each pair to be a single object and every person who sticks gum below his/her chair also as a single object. If there are $r$ such pairs (call them double objects), then there would $10-2r$ single objects. Hece there would be total $10-2r+r=10-r$ total objects which can be arranged in $^{10-r}C_{r}$ ways. Hence in total there are:

\sum_{r=0}^{5}^{10-r}C_{r}=\boxed{89} ways.

Fibonacci series! Try it with two seats, then three, then four seats.

For each seat you add, you have the same set as before ( he puts it on his own seat - the n-1 set) plus a smaller set ( in which he puts his gum on his neighbors seat).

Since the second case "takes away" the last two seats ( the have to swap), it's the same as the ( n-2) set.