wcn@cs.brown.edu (Wen-Chun Ni) (12/03/90)
I gave an erroneous answer k^n/k! for the partitioning problem since the solution presumed the existence of the "empty groups" even we just think them unlabelled. So the answer k^n applies only when we make the group labelled. The answer sum(for i=1 to k) S(n,i) (the Bell number )is correct in the unlabelled case. I thank B. S. Yee at CMU for pointing out the error. Generally, we just put the following ways of enumerations: members labelled and groups labelled : k^n members labelled and groups unlabelled: sum (for i=1 to k) S(n,i) members unlabelled and groups labelled: {n+k-1 choose k} members unlabelled and groups labelled: sum (for i=1 to k) p(n,i). The quantity p(n,i) is the number of ways partitioning an integer into i parts (eg. 4=4=1+3=2+2=1+1+2=1+1+1+1). If we want something like "monotonicity," "injective," or "surjective," please refer to "Enumerative Combinatorics" by Richard Stanley.