Consider $\sum k(k-1)(k-2)$ $n\choose k$ = $n(n-1)(n-2)$ $n-3 \choose 3$ for k >= 0 n >=3
I had initially thought the right side counted the ways to select three distinct objects from n and then choose an additional three arbitrarily from the remaining objects but I was unsure on how the LHS summation fit into the picture. Does it make sense to choose distinguished objects for all k? I feel like there is another identity that ties into this, where the sum $\sum$ $n \choose k$ over all k is just the same as choosing from the largest element. Does anyone recall whether something like that exists?