A more important and meaningful question would be to ask if there is a reliable mathematical approach. Sometimes, but not always, a formula is the result of such an approach. I wouldn’t expect there to be a single formula, but rather a set of formulae to be used logically and systematically by thinking the problem through.
The first formula is one which determines the number of combinations given a set of items to choose from, and then selecting a certain number of these items.
Suppose the list, or set X, of all items has size n and the two selection lists, List 1 and List 2, have equal size m. It’s clear that n≥2m. Then we come to the number of mutually exclusive items, size p≥2, for a set E of such items.
Start with nCr which is the number of combinations of r items chosen from a set of n objects. The calculation is n(n-1)(n-2)...(n-r+1)/(r(r-1)...(2)(1)). This can be written as a formula: n!/(r!(n-r)!). For example, the number of combinations of selecting 6 objects from a set of 13 is 13!/(6!7!)=13×12×11×10×9×8/(6×5×4×3×2×1)=1716.
To form two lists with NO exclusions (p=0), there are (nCm)((n-m)Cm) combinations. For example with n=13 and m=6 this is 13C6×7C6=12,012. For n=16 and m=6 it’s 16C6×10C6=1,681,680.
So one approach could be to calculate the no-exclusions figure then subtract the exclusion combinations. So next we look at the exclusions. We need to produce a set of exclusive combinations from the simple list of items that can’t appear together in the same list. So we remove these items from the main set, leaving n-p items. Now let’s use some set notation. X={x1 x2 ... xn}, and E={e1 e2 ... ep}, where these e elements also reside in X. Now we define set Y=X-E, which has size n-p, and consists of all items in X that are not also in E.
Next we start to build the exclusion combination set, S. We take a pair of elements from E, say e1 and e2, and then we combine these with m-2 elements of Y. We do this for all possible pairs of elements in E and m-2 elements from Y. We can work out pC2 × (n-p)C(m-2), where pC2 applies to E and (n-p)C(m-2) applies to Y. Then we calculate pC3 × (n-p)C(m-3) for sets of 3 elements in E, and so on up to pCp (all of E)=1 and (n-p)C(m-p).
For example: n=13, m=6, p=3 gives us 3C2×10C4+1×10C3=630+120=750. So the size of S is 750 in this case.
For n=16, m=6, p=4 we get 4C2×12C4+4C3×12C3+12C2=3,916 as the size of S.
Let’s take a simple example and apply what we have so far.
A SIMPLE CASE
X={a b c d e f g}, E={b d}, so n=7 and p=2. S={abd bcd bde bdf bdg} and the size of S=pC2×(n-p)C(m-2)=2C2×5C1=5 (2C2=1); Y=X-E={a c e f g} has a size of n-p=5. Lists 1 and 2 are to have 3 items each, so m=3.
Set of lists: Q is the set of all possibilities for List 1. The combinations are referred to as combos.
Q={abc abd abe abf abg acd ace acf acg ade adf adg aef aeg afg bcd bce bcf bcg bde bdf bdg bef beg bfg cde cdf cdg cef ceg cfg def deg dfg efg} (7C3=35 combos)
Next, we remove excluded combinations (the ones marked with strikethrough) and redefine Q as Q-S➝Q:
Q={abc abe abf abg acd ace acf acg ade adf adg aef aeg afg bce bcf bcg bef beg bfg cde cdf cdg cef ceg cfg def deg dfg efg} (30 combos)
For each combo in Q we need to work out a set of List 2 combos. First we ignore excluded combos, but mark them with strikethrough:
abc: {def deg dfg efg} adg: {bce bcf bef cef} cde: {abf abg afg bfg}
abe: {cdf cdg cfg dfg} aef: {bcd bcg bdg cdg} cdf: {abe abg aeg beg}
abf: {cde cdg ceg deg} aeg: {bcd bcf bdf cdf} cdg: {abe abf aef bef}
abg: {cde cdf cef def} afg: {bcd bce bde cde} cef: {abd abg adg bdg}
acd: {bef beg bfg efg} bce: {adf adg afg dfg} ceg: {abd abf adf bdf}
ace: {bdf bdg bfg dfg} bcf: {ade adg aeg deg} cfg: {abd abe ade bde}
acf: {bde bdg beg deg} bcg: {ade adf aef def} def: {abc abg acg bcg}
acg: {bde bdf beg def} bef: {acd acg adg cdg} deg: {abc abf acf bcf}
ade: {bcf bcg bfg cfg} beg: {acd acf adf cdf} dfg: {abc abe ace bce}
adf: {bce bcg beg ceg} bfg: {acd ace ade cde} efg: {abc abd acd bcd}
When the 20 marked ones are removed we are left with 100 combinations of List 1 with List 2. Half of these are duplicates, in that List 1 and List 2 are simply interchanged; for example, List 1=abc and List 2=def may be regarded as the same as List 1=def and List 2=abc. If this is the way combinations are to be measured then there are only 50 unique combinations.
I hope this gives you some idea of how to approach the problem.