Question: Use combinatorial reasoning to prove the identity:

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

My reasoning for the LHS side is that we have a set of $n$ objects, with three special ones. We want to have all sets of size $k$ which contain at least one of these special objects. So we take all possible $k-combinations$, then we subtract the $k-combinations$ without any of these three special ones. 

However, I am stuck about how to apply this same situation to the RHS.

Answer: 
HINT: Let the set of $n$ objects be specifically the set $[n]=\{1,2,\ldots,n\}$, and let the three special objects be the numbers $n,n-1$, and $n-2$. For each of the special numbers count the number of $k$-element subsets of $[n]$ that have that special element as largest member.
To generalize Brian M. Scott's answer, try proving that ${n \choose k}={{n-1}\choose {k-1}} + {{n-1}\choose {k}}$. Like Brian's answer, consider the set $[n]=\{1,2...n\}$ but then just consider $n$ and how many size $k$ subsets of $[n]$ contain $n$ and how many don't.