I am trying to prove algebraically and using combinatorics that $$\binom{n}{2} - 1 = 2(n-2) +\binom{n-2}{2}$$ where $n\geq2$
I had no issues doing the algebraic proof, but I am struggling a lot with the combinatorial proof. I understand the left hand side is showing how to pick the subsets of size 2 out of n elements minus one (for the empty set?), but I am unsure how to explain the right hand side. Is there a direct and/or simple approach that perhaps I am overlooking?
Here is a combinatorial approach with a story. Consider there is a class of $n$ students. \begin{equation} C = \{1,2,...,n\} \end{equation} Let's say we will pick $2$ students from the class, but we are not allowed to pick the students $1$ and $2$ together, say they don't like each other. Then LHS and RHS can be given by two different counting approaches in which we count all the possible choices of two students in the desired way as follows:
LHS: We count in how many ways we can pick $2$ students from the class and exclude the set $\{1,2\}$ since they don't want to be chosen together. That will give us $\binom{n}{2}-1$.
RHS: We exclude those two at the beginning and count in how many ways we can pick $2$ students from the rest of the class ($n-2$ students) and then we can choose $1$ student from the set $\{1,2\}$ and $1$ from the rest $n-2$ student. That will give us $\binom{n-2}{2} +\binom{2}{1}\binom{n-2}{1}=2(n-2)+\binom{n-2}{2}$.