In how many ways can we color exactly $k$ vertices of an $n$-gon in red such that any $2$ consecutive vertices are not both red. (Vertices are considered to be labeled)
1. **Define the problem for a straight line:**
   Let \( f(n, k) \) denote the number of ways to color exactly \( k \) points red on a straight line of \( n \) points such that no two consecutive points are both red.

2. **Establish the recurrence relation:**
   - If the \( n \)-th point is red, then the \((n-1)\)-th point cannot be red. The remaining \( n-2 \) points must have \( k-1 \) red points, which can be done in \( f(n-2, k-1) \) ways.
   - If the \( n \)-th point is not red, then the remaining \( n-1 \) points must have \( k \) red points, which can be done in \( f(n-1, k) \) ways.
   - Therefore, the recurrence relation is:
     \[
     f(n, k) = f(n-1, k) + f(n-2, k-1)
     \]

3. **Prove the closed form by induction:**
   - Base case: For \( k = 0 \), \( f(n, 0) = 1 \) if \( n \geq 0 \) (since there is exactly one way to color zero points red).
   - Inductive step: Assume \( f(n, k) = \binom{n-k+1}{k} \) holds for all \( n \leq m \). We need to show it holds for \( n = m+1 \).
     \[
     f(m+1, k) = f(m, k) + f(m-1, k-1)
     \]
     By the inductive hypothesis:
     \[
     f(m, k) = \binom{m-k+1}{k}, \quad f(m-1, k-1) = \binom{m-k+2}{k-1}
     \]
     Using the Pascal's identity:
     \[
     \binom{m-k+1}{k} + \binom{m-k+2}{k-1} = \binom{m-k+2}{k}
     \]
     Therefore:
     \[
     f(m+1, k) = \binom{m-k+2}{k}
     \]
     This completes the induction.

4. **Return to the original problem for an \( n \)-gon:**
   - Fix a vertex \( P \). If \( P \) is red, the two adjacent vertices cannot be red. The remaining \( n-3 \) vertices must have \( k-1 \) red vertices, which can be done in \( f(n-3, k-1) \) ways.
   - If \( P \) is not red, the remaining \( n-1 \) vertices must have \( k \) red vertices, which can be done in \( f(n-1, k) \) ways.
   - Therefore, the total number of ways is:
     \[
     f(n-3, k-1) + f(n-1, k)
     \]

5. **Substitute the closed form:**
   \[
   f(n-3, k-1) = \binom{(n-3)-(k-1)+1}{k-1} = \binom{n-k-1}{k-1}
   \]
   \[
   f(n-1, k) = \binom{(n-1)-k+1}{k} = \binom{n-k}{k}
   \]
   Therefore, the required answer is:
   \[
   \binom{n-k-1}{k-1} + \binom{n-k}{k}
   \]

The final answer is \( \boxed{ \binom{n-k-1}{k-1} + \binom{n-k}{k} } \)