Types of flops, in relation of held hand

Question

I'm having difficulties on matching the expected total number of flops $50\choose3$ $= 19600$ by splitting them in categories in relation to the held hand, let's say for example it is AS (Ace of spades), KD (King of diamonds), so this categories can be defined:

(Lets say A is Spades, B is Diamonds, X is one of the others left, and Y the last one)

1. Flop of 3 cards suited with A: $12\choose3$ $= 220$ flops in this category.
2. Flop of 3 cards suited with B: $12\choose3$ $= 220$ flops in this category.
3. Flop of 2 cards suited with A, one with B: $(12·11·12)/3!$ $= 264$ flops in this category.
4. Flop of 2 cards suited with B, one with A: $(12·11·12)/3!$ $= 264$ flops in this category.
5. Flop of 2 cards suited with A, one non suited with hand suits: $(12·11·26)/3!$ $= 572$ flops in this category.
6. Flop of 2 cards suited with B, one non suited with hand suits: $(12·11·26)/3!$ $= 572$ flops in this category.
7. Flop of 1 card suited with A, 1 suited with B, 1 non-suited: $(12·12·26)/3!$ $= 624$ flops in this category.
8. Flop of 1 card suited with A, 2 non-suited of the same suit: $(12·26·12)/3!$ $= 624$ flops in this category.
9. Flop of 1 card suited with B, 2 non-suited of the same suit: $(12·26·12)/3!$ $= 624$ flops in this category.
10. Flop of 1 card suited with A, 2 non-suited, not of the same suit: $(12·26·13)/3!$ $= 676$ flops in this category.
11. Flop of 1 card suited with B, 2 non-suited, not of the same suit: $(12·26·13)/3!$ $= 676$ flops in this category.
12. Flop of 3 non-suited, all of same suit: $(26·12·11)/3!$ $= 572$ flops in this category.
13. Flop of 3 non-suited, 2 of them same suit: $(26·12·13)/3!$ $= 676$ flops in this category.

The sum of flops in this categories (and I guess I'm missing none) accounts for a total of $6584 \neq 19600 $ (total flops possible).

I reviewed the thing a number of times, spent time on it, and I cannot find where my reasoning is flawed, so I'm feeling like a dork, or maybe I found a hole in an centuries old math area (I guess beatings are better taken with humour).

If anyone is so kind to point the thing out, I'm grateful.

Best regards.

Answer 1

You’re dividing everything by $3!$; in most cases that’s wrong; that factor only arises in $\binom{12}3=\frac{12!}{9!3!}$.

I’ll go through the list one by one:

1. correct
2. correct
3. This should be $\binom{12}2\binom{12}1=792$.
4. As in 3.
5. This should be $\binom{12}2\binom{26}1=1716$.
6. As in 5.
7. This should be $\binom{12}1\binom{12}1\binom{26}1=3744$.
8. This should be $\binom{12}1\binom21\binom{13}2=1872$ (with $\binom21$ for choosing the non-hand suit).
9. As in 8.
10. This should be $\binom{12}1\binom{13}1\binom{13}1=2028$.
11. As in 10.
12. This should be $\binom21\binom{13}3=572$ (with $\binom21$ for choosing the non-hand suit).
13. This should be $\binom21\binom{13}2\binom{13}1=2028$ (with $\binom21$ for choosing the non-hand suit with two cards).

The total is $2\cdot220+2\cdot792+2\cdot1716+3744+2\cdot1872+2\cdot2028+572+2028=19600$, as expected.