TITLE: People sitting in a circle chewing gum
QUESTION [6 upvotes]: Ten people are sitting in a circle of ten chairs, chewing gum. Each person spits out his or her gum and places it either under his or her own chair or under an immediately adjacent chair. How many ways can this happen such that every chair ends up with exactly one piece of gum under it?

I tried creating a chart of small values, obtaining for n people, when n = {1, 2, 3, 4, 5}, $f_n$ = {1, 2, 6, 9, 13}
At first I thought tribonacci, as soon as I saw $f_4$ but that notion vanished as I calculated $f_5$
Can someone provide a clear answer and or solution for $f_{10}$? Thanks.

REPLY [3 votes]: We can systematically decompose the problem top-down. Let $f(n)$ be the direction in which the $n$-th person (with indices wrapping round the ring) does his disgusting thing, $-1,0,1$ for left, centre and right respectively. Note that the number of ways is equivalent to the number of possible $f$ that satisfies the corresponding conditions, as long as $n \ge 3$. The obvious way is to take out one person. If $f(0) = 0$, then we need to know the number of ways for $(n-1)$ people such that $f(0) \ne 1$ and $f(1) \ne -1$. If $f(0) = 1$, then either $f(1) = -1$ or $f(-1) = 1$. In the first case we need to know the number of ways for $(n-2)$ people such that $f(0) \ne 1$ and $f(1) \ne -1$. In the second case we need to know the number of ways for $(n-1)$ people such that $f(0) = 1$ and $f(1) \ne -1$. I will leave you to verify the bijection in each of these reductions. These reductions tell us to define the following:
$\def\nn{\mathbb{N}}$
Let $a(n) = \#(\text{ways for $n$ people such that $f(0) \ne 1$ and $f(1) \ne -1$})$.
Let $b(n) = \#(\text{ways for $n$ people such that $f(0) = 1$ and $f(1) \ne -1$})$.
Then $a(n+2) = a(n+1) + a(n)$ for any $n \in \nn^+$. [Either $f(1) = 0$ or $(f(1),f(2)) = (1,-1)$.]
Also $b(n+1) = b(n)$ for any $n \in \nn^+$. [$f(-1) = 1$.]
You can check that $a(1) = 1$ and $a(2) = 2$, and hence $a(n) = F_{n+1}$ for any $n \in \nn^+$ where $F_n$ is the $n$-th Fibonacci number. You can also check that $b(1) = 1$, and hence $b(n) = 1$ for any $n \in \nn^+$.
Therefore the total number of ways for $n$ people is
  $a(n-1) + 2 a(n-2) + 2 b(n)$
  $\ = F_n + 2 F_{n-1} + 2$
  $\ = F_{n+1} + F_{n-1} + 2$.
Notes
This method gives exactly the same answer as Henry's method, but is more general and can be applied to other more complicated problems without requiring any insight.