Counting isomorphic combinations Assume a simplified poker deck (to cut down the # of combinations):


*

*4 suits (h, c, d, s)

*5 ranks (A, K, Q, 7, 2)

*20 cards


Assume a basic 2-player setup:


*

*Hand 1 (2 cards)

*Hand 2 (2 cards)

*Flop (3 cards)


There are 16,279,200 combinations of the above setup: choose(20, 2) * choose(18, 2) * choose(16, 3).
Take the following combination:


*

*Hand 1 = Ad, As

*Hand 2 = Kc, Kh

*Flop = Qd, 7s, 2c


Question: what is the most efficient way to calculate the number of combinations (within the 16,279,200) that are suit-isomorphic with this setup?
For example, the following setup is suit-isomorphic:


*

*Hand 1 = As, Ac

*Hand 2 = Kh, Kd

*Flop = Qs, 7c, 2h


while this example is not:


*

*Hand 1 = Ad, As

*Hand 2 = Kc, Kh

*Flop = Qc, 7s, 2d


nor is this example:


*

*Hand 1 = Kc, Kh

*Hand 2 = Ad, As

*Flop = Qd, 7s, 2c


===Edit ... explaining what I mean by suit isomorphism===
The only isomorphic transformation allowed is one that shifts suits, not ranks. The first example failed example isn't "suit" isomorphic because f(d) isn't a one-to-one map: on the Ace in Hand 1, f(d) = d. On the Queen in the flop, f(d) = c. The second example failed because the ranks within Hand 1 and Hand 2 were swapped.
===Edit ... a more general approach to the question===
The solution shouldn't rely on a specific structure of the problem. For example, in the example I gave, the solution is 24 (i.e. the permutations of the 4 suit symbols). But what happens in the following cases.


*

*I might start eliminating cards. For example, I might pull the 2c out of the deck.

*I might not use all 4 symbols.

 A: You are asking for an orbit of a group action, where the group is the group of all permutations of the suits, and the action is the action on the setups where a permutation acts on each card individually.  The orbit can just be constructed by taking the 24 possibilities and seeing which ones are "allowed" and which are different.  24 is a tiny number.
Sometimes though permuting the suits may not actually change the setup.  Consider the example:


*

*Hand 1: 3♣, 3♠

*Hand 2: 3♢, 3♡

*Flop: 2♡, 4♡, 5♡


Then the permutation that switches ♣ and ♠ but leaves ♢ and ♡ alone actually ends up leaving the setup alone: hand 1 switches the order of the two cards, but that doesn't change the hand; hand 2 and the flop stay exactly the same.
Assuming you don't throw any cards out, the number of isomorphic setups is always a divisor of 24, namely the index of the "stabilizer" of the setup (see the "orbit-stabilizer theorem").
If you start throwing out cards, then basically your problem has no structure.  This is not a big problem, because 24 is a tiny number.  Just try all 24 and see which are "allowed".
