I have encountered a counting problem which involves counting all ordered partitions, where each partition can also have one of n labels.
For example: Given a set of two elements {A,B} with a single label L, there are two partitions, {L{A,B}} and {L{A},L{B}}, but there are four ordered partitions, (L{A,B}), (L{B,A}), (L{A},L{B}), and (L{B},L{A}).
If we have 2 possible labels X and Y, we now have the following possible labeled ordered partitions: X:{A,B} Y:{A,B} X:{B,A} Y:{B,A} X{A},X{B} X{A},Y{B} Y{A},X{B} Y{A},Y{B} X{B},X{A} X{B},Y{A} Y{B},X{A} Y{B},Y{A}
Are there closed formulas for this problem? Is this problem well studied in combinatorics? Are asymptotic formulas known?
I am happy for any advice or reference I can get.
I think that what you have in mind is not an ordered partition, because your subsets have interior orders; it's really an ordered list of labelled disjoint tuples on the given set. So you can express the list as a permutation of the set with a label added next to the last element of each tuple.
Suppose the given set has ten elements, $\{0,1,\ldots 9\}$, and we have decided to use four labels. Write down one of the $10!$ permutations of this set. Now, after each element, either write a label, or a comma. The comma is effectively a fifth label. The last element, however, must have one of the four real labels. Thus the number of possibilities is $10!\cdot5^9\cdot4$.
Similarly, the number of possibilities on a set of $n$ elements and a set of $L$ labels is $$n!L(1+L)^{n-1}$$