How to prove a combinatoric statement? 
From Number 10B with PICTURE. Suppose there are n plates equally spaced around a circular table. Ross wishes to
  place an identical gift on each of k plates, so that no two neighbouring plates have
  gifts. Let f(n, k) represent the number of ways in which he can place the gifts. For
  example f(6, 3) = 2, as shown below. Prove that f(n, k) = f(n−1, k) +f(n−2, k −1) for all integers n ≥ 3 and k ≥ 2.

Trying combinatoric, and need help with induction
I was thinking of a 'proof' involving combinatorics. Here is an idea.
Consider two cases:
Case 1: Plate at table one. Then going in front, there are $n-1$ tables left, but the next one cannot have a plate, so it must start at the $n-2$nd plate. There are $k-1$ plates left, the number of arrangements should equal: $f(n-2, k-1)$. Likewise the argument works for Case 2, with no plate at table one.
But is this a strong enough proof?
How can I strengthen a proof?
 A: I think you're confusing tables with plates and plates with gifts, though I agree with your proof (mostly). It's $f(n-2,k-1)$ because you have to exclude both plates to each side of the plate that you start with. I would say something like, "Fix a plate", since they technically don't say that the plates are distinct, so you might not be able to say that there's a plate number 1. You should probably also flesh out the other case just to be sure you can put it in words, even if its obvious.
A: We will fix the plates in their position and number them $1,2,3,\ldots$ in clockwise order.
Let $A_{n,k}$ denote the set of valid arrangements of $k$ gifts on $n$ plates. Then $f(n,k) = \vert A_{n,k}\vert$ and we will prove the claim by setting up a bijection (graphically) from $A_{n-2,k-1}\cup A_{n-1,k}$ to $A_{n,k}$, which involves adding plate number $n$ and, if required, $n-1$.
In the following table, "$\bullet$" denotes a plate with a gift and "$\circ$" denotes a plate without a gift.
The following mapping depends on the state of the first and last plates, which are neighbours of course, of sets $A_{n-2,k-1}$ and $A_{n-1,k}$.
$$
\begin{array}{ccccccccc}
\text{Plate No.: } & 1 & n-2 &  & \quad 1\quad & \quad \color{red}n \quad & \color{red}{n-1} & n-2 & \\
A_{n-2,k-1} & \bullet & \circ & \qquad \mapsto \qquad & \bullet & \color{red}\circ & \color{red}\bullet & \circ & A_{n,k} \\
A_{n-2,k-1} & \circ & \bullet & \mapsto & \circ & \color{red}\bullet & \color{red}\circ & \bullet & A_{n,k} \\
A_{n-2,k-1} & \circ & \circ & \mapsto & \circ & \color{red}\bullet & \color{red}\circ & \circ & A_{n,k} \\
& \\
\text{Plate No.: } & 1 & n-1 & & \quad 1\quad & \quad \color{red}n \quad & n-1 & & \\
A_{n-1,k} & \bullet & \circ & \mapsto & \bullet & \color{red}\circ & \circ & & A_{n,k}\\
A_{n-1,k} & \circ & \bullet & \mapsto & \circ & \color{red}\circ & \bullet & & A_{n,k}\\
A_{n-1,k} & \circ & \circ & \mapsto & \circ & \color{red}\circ & \circ & & A_{n,k}\\
\end{array}
$$
Note that for each of the three sets, we have included all members exactly once, hence the bijection is established and the proof is complete.
