Combinatorial Proof for $\binom{n}{k}-\binom{n-3}{k} = \binom{n-1}{k-1} + \binom{n-2}{k-1} + \binom{n-3}{k-1}$

Question

I am struggling to see how to go about a combinatorial proof of the following:

$$\binom{n}{k}-\binom{n-3}{k} = \binom{n-1}{k-1} + \binom{n-2}{k-1} + \binom{n-3}{k-1}$$

I have an algebraic proof of this, but I'm not sure how to view this in order to prove it combinatorially. Any help with this would be greatly appreciated.