My solution: Look at one wi. If it is alone in the set there are S(n-1,k-1) ways of arranging the rest and the weight of wi is 1. Or wi isn't alone. Then first partition the rest there are S(n-1,k) ways of doing that. Then we place wi into one of those set and sum the sizes of these new sets. Together (sum wi)(S(n-1,k-1) + (n-1+k)S(n-1,k)).