For example,
Needs["Combinatorica`"]
n = 4;
partitions = SetPartitions[Range[n]];
Table[Print[partitions[[i]]], {i, 1, BellB[n]}]
generates
{{1,2,3,4}}
{{1},{2,3,4}}
{{1,2},{3,4}}
{{1,3,4},{2}}
{{1,2,3},{4}}
{{1,4},{2,3}}
{{1,2,4},{3}}
{{1,3},{2,4}}
{{1},{2},{3,4}}
{{1},{2,3},{4}}
{{1},{2,4},{3}}
{{1,2},{3},{4}}
{{1,3},{2},{4}}
{{1,4},{2},{3}}
{{1},{2},{3},{4}}
(and a bunch of Nulls !)
I notice that the coarsest ordering of these partitions is in increasing order of the number of parts, $k$. Also the partitions are written such that the elements of each part are in increasing order, and the parts are in increasing order of their least (hence also first appearing) element. I notice that for each $k$ (the number of parts), the partitions having $\{1\}$ as a singleton set are listed first, and that this rule seems to be applied recursively to the partition of the remaining set.
@Mr.Wizard gives some nice code here, for generating the same partitions, that happens to order the partitions differently.
Print[]isNull, which gets passed toTable[]. TryScan[Print, partitions]if you wish for a one-by-one printout. As for the ordering, IIRC this is in lexicographic order. – J. M.'s missing motivation Sep 25 '18 at 20:30