Combinatorics question-- suited combinations?
So, I know that there are 4 combinations of any 2 specific suited cards. For example, 4 combos of QKs, 4 combos of AKs, etc. Where I run into problems with finding the number of combinations of suited cards in a villain's range are situations where villain plays suited kings, queens, or God forbid, any two suited cards. If villain plays only suited aces, I think that he would have 48 combos of any suited ace, and only 12 that match the suit in question on the board, unless there are 2 double suits by the turn in which case it would be 24, right? What if a looser player, whether it be from late position or just a bad player from EP, opens any suited ace, king, or queen? So, removing the Ace of that suit that we already accounted for when counting suited aces, would there be 44 combos of suited kings, with 11 of them matching the suit in question on the board or 22 on double suited turn boards? Then, when estimating suited queens, and removing the previously accounted for suited aces and kings, would there be 40 combos with 10 that match the board's suit? I'm really bad with figuring this out when it comes to removing the previously accounted for suited hands. Can anyone give me a formula for this? If not a formula, even just answering how many combos of suited cards a villain could have if he plays only suited aces, suited aces+kings, suited aces, kings, and queens, and the dreaded any two suited cards would help me out a lot. Maybe there's an easy way to figure this out that I can't put my finger on. Please help!