Problem: Compute $ \sum_{k=0}^n k^2\binom{n}{k} $ Solution: Imagine you want to select color pallets from $n$ colors and you will have a primary and secondary color along with however many accent colors we'd like. The primary color may be the same as the secondary color (it's double counting so anything goes...). The set of all such color pallets is exactly counted as the above: For each $k$ such that $0 \leq k \leq n$, we may select $k$ colors from $n$ total colors, and then we choose, from among $k$, exactly one color to be the primary, and exactly one to be the secondary. Now, consider the following way of counting these pallets. Simply choose your primary and secondary from among the $n$. and then tack on as many or as little accent colors as we'd like. When the primary is the same as the secondary, there are $n {2}^{n-1}$ such pallets, and when the primary and secondary are unique, there are exactly $n(n-1)2^{n-2}$ such pallets, so in total, there are $n(n+1) \cdot 2^{n-2}$ pallets!