Simplifying $\binom{n-k}{k} + \binom{n-k-1}{k-1}$

Question

I am trying to find the formula for the number $g(n,k)$ which is the number of $k$-subsets of $[n]$ where no two elements are consecutive numbers in $[n]$ arranged in a circle.

I derived the following recursion $$f(n,k) = f(n-1,k) + f(n-2, k-1) $$ where $f(n,k)$ is the number of ways to take $k$=subsets from $[n]$

and also the formula of $f(n,k)$ as following $$f(n,k) = \binom{n-k+1}{k}$$

Also,

I derived following relation

$$g(n,k) = f(n-1,k) + f(n-3, k-1)$$

so that g(n,k) is just represented as $$\binom{n-k}{k} + \binom{n-k-1}{k-1}$$

Is there ways to merge this two binomials given as summation?

Answer 1

A further simplification could be $$\binom{n-k}{k} + \binom{n-k-1}{k-1}=\binom{n-k}{k} + \frac{k}{n-k}\binom{n-k}{k}=\frac{n}{n-k}\binom{n-k}{k}.$$

Answer 2

Well, you can always do something like $$ \binom{n-k}{k} = \frac{(n-k)!}{k! (n-2k)!} = \frac{n-k}{k} \frac{(n-k-1)!}{(k-1)!(n-2k)!} = \frac{n-k}{k} \binom{n-k-1}{n-1}, $$ and then merge the two coefficients.

Answer 3

Or: $$\binom{n-k}{k} + \binom{n-k-1}{k-1}=\frac{n}{k}\binom{n-k-1}{k-1}=\frac{n}{k}\cdot \frac{(n-k-1)!}{(k-1)!(n-2k)!} \hspace{4.7cm} =\frac{n(n-k)!}{(n-k)\cdot k!(n-2k)!}=\frac{n}{n-k}\binom{n-k}{k}.$$