Subset of Permutations of distinct symbols, with positional count constraint/ I am working a personal project, and I need to derive subsets of permutations of unique elements.
For example:  I have a set of 5 unique elements {A, B, C, D, E}.   There are 60 permutations tuples of size 3 of this: {A,B,C}, {A,B,D}...{E,C,D}, {E,D,C}.  What I want is to select tuples from this list such that the count of each symbol in the same position across all tuples is constrained as 0 < A# < B# < C# < D# < E# the count of each symbol is the same in each position.
E.g.
{A, B, C}
{A, B, D}
⋮
{E, C, D}
{E, D, C}

For the full set of tuples, there are 12 of each symbol in each position within the tuple, or specifically for what is explicitly written above there are quantity A2 and E2 in position 0, B2 C1 D1 in position 1, and C2 and D2 in position 2, so this does not meet the criteria.
With the example of 5 permute 3, as there are only 60 permutation tuples I was able to create such a desired subset by brute force, finding a collection of 20 permutation tuples resulting in counts of A2, B3, C4, D5, E6 in each position.
My question is:   Is there a known algorithmic way of selecting subsets with this type of criteria, and is there a predictive formula to calculate how many valid solutions can be derived?  I will need this to be for quantities of symbols up to 15 and up to tuple size 5 so brute force is sub-optimal.
For completeness, the solution subset I found for 5 symbols permut 3 is the following set:
AEC
AED
BCE
BDE
BED
CBE
CDB
CDE
CED
DBC
DBE
DCE
DEB
DEC
EAC
EAD
ECA
ECD
EDA
EDB

Edit:  Using logic inspired by Ross Millikan, I verified that 4 permut 3 has no solution, as the permut domain is too small.  I did find via brute force a solution for 6 permut 3: A3,B4,C5,D6,E7,F8 for a total of 33 tuples:
ABC
ABE
ACD
BCF
BDF
BEF
BFE
CAD
CED
CEF
CFD
CFE
DCE
DCF
DEC
DEF
DFC
DFE
ECF
EDB
EDC
EDF
EFB
EFC
EFD
FAD
FAE
FBA
FBE
FDA
FDB
FEA
FEB

 A: Not an answer, but I hope helpful and too long for a comment.
There are a number of constraints that you can use to guide the search.  The big one is that the sum of the counts in each position must match the number of tuples.  For your example, you have $20$ tuples and $2+3+4+5+6=20$  Among the $60$ permutations you have $E$ at each slot in $12$ of them.  If you were to demand $E\#=12$ you would need $36$ permutations to satisfy that because there are $12\ E$s in each position, which is all the permutations that include $E$.  There are $24$ non-E slots which are equally distributed among the other letters, violating your constraint that there be less $A$s than $B$s and so on.
The number of permutations is the sum of the counts because you have one letter in each position of each permutation.  If you demanded $E\#=11$ you would need at least $33$ permutations (more if some do not contain $E$ at all).  The sum of the other four counts would be at least $22$.  That means $D\#$ is at least $7$, so there are at least four $D$s in one position matched with $E$s in the same position.  However there are only three choices for the third position, so this will not work.  The same argument works for $E\#=10$.  You have a hope with $E\#=9, 27$ total permutations with $D\#=6, C\#=5, B\#=4, A\#=3$  Three of the $D$s in each column go with $E$s in each of the other two.  That gives you $18$ permutations including $DE$ and  one of each of the other letters and uses two of the other letters in each column.  If we focus on the ones with $E$ in the first column we have
E,D,(A,B,C)
E,(A,B,C),D
for six and we need three more which can only be
EAC
ECB
EBA
and we have an equal number of $A,B,C$ so this fails.
I don't think it is easy
