Number of combinations when swapping two sets with some elements in both This question is closely related to this other one: I have two sets and I want to know the number of possibilities I can do with the elements of these two sets. 
As a possibility I mean changing elements ( 1 or several) from one set with elements from another set, but elements in a set cannot be repeated. 
However in contrast with the other question, I have some element that are in both sets, and remember that I don't want the same element repeated in the same set:
With these pair of sets set1: {A, B, C, D, E, F, G}; set2: {A, D, H}, I would say that there are 5 possibilities:  5 elements from set1 that can be swapped with one element of set2:
Possibilities:
set1:{A, D, H, C, E, F, G}; set2: {A, D, B}
set1:{A, D, H, B, E, F, G}; set2: {A, D, C}
set1:{A, D, H, B, C, F, G}; set2: {A, D, E}
set1:{A, D, H, B, C, E, G}; set2: {A, D, F}
set1:{A, D, H, B, C, E, F}; set2: {A, D, G}

If element A or D are swapped, they will be repeated in one set (unless they are changed with themselves in which case we haven't done anything), and if element H is not in set1 it wouldn't have changed any element from set2 to set1.
But with these other pair, set1: {A, B, C, D, E, F, G}; set3: {A, H, I} I think that there are 12 possibilities with just changing one element between the sets (6*2) and then 15 if we change 2 elements from one set to the other ${6 \choose 4}$, in total 27.
I can't find a rule to generalize to different settings (having more shared elements, different size of sets, ...). Is there any mathematical/statistical rule to count this?

It is different from Cartesian product because I don't want to know the number of sets that can be created taking one element from each set. 
 A: You seem to ask the following: given two sets $A$ and $B$ (in order), in how many ways can a new pair of sets of the same sizes as the original sets be created by removing one or more elements from $B$ and the same number of elements from $A$ and putting each collection back into the other set?
I hope it's obvious that we don't need to keep track of elements common to $A$ and $B$ (because swapping them does nothing).  Thus, we may remove all common elements from each set, reducing the question to the case where $A$ and $B$ are disjoint.
Notice that the result consists of an ordered pair of disjoint subsets of the union of $A$ and $B$ and is determined (say) by the elements of the first set, of which there are $|A|$.  Consequently, the total number of swaps--including doing nothing, which yields the original pair of sets--is in one-to-one correspondence with the subsets of size $|A|$ in the union $A\cup B$, which has size $|A|+|B|$.  Subtracting $1$ to account for the original configuration gives
$$\binom{|A|+|B|}{|A|}-1$$
as the answer.  In general, if you want to be explicit about removing the common elements $A\cap B,$ the formula is
$$\binom{|A|+|B|-2|A\cap B|}{|A|-|A\cap B|}-1.$$
This formula does indeed count combinations, in sense understood in combinatorics and probability theory--but they are combinations within the derived set $A\, \nabla\, B,$ the symmetric difference of $A$ and $B.$

In the second example of the question where the sets are $\{a,b,c,d,e,f,g\}$ and $\{a,h,i\},$ we have $|A|=7,$ $|B|=3,$ and $|A\cap B|=1,$ yielding
$$\binom{7+3-2(1)}{7-1}-1 = \binom{8}{6}-1 = 28-1 = 27.$$
