Problem: Prove that $\binom{n \choose 2}{2}=3\binom{n}{4}+3\binom{n}{3}$ Solution: Consider the set of all pairings (order doesn't matter) of distinct edges in a graph $G$ on $n$ vertices. Indeed, there are ${n \choose 2}$ edges (choosing two verticies specifies an edge) and hence there are ${{n \choose 2} \choose 2}$ such pairings. On the other hand, we can count this by casing on how many distinct verticies our edges occupy. say the two edges occupy $4$ unique verticies. Then, there are ${n \choose 4}$ ways to choose these vertices and exactly $3$ unique ways to draw edges between them so that no vertices have degree $2$. Similarly, if two edges share a common vertex, then, there are ${n \choose 3}$ ways to pick three vertices from $n$ and then $3$ ways to draw two edges between them. Hence, the conclusion follows.