I'm trying to calculate the size of non-equivalent poker hands with a given suit configuration (additional context here). from a related question:
Lets write $[x_1, y_1][x_2, y_2]\ldots$ for number configuration with $y_1$ suits of $x_1$ cards, $y_2$ suits of $x_2$ cards and so on. We have $[x_1, y_1]\ldots = \prod_i [x_i, y_i]$ - suits with different number of cards are indepdent.
Now, for single suit of $x$ cards, we have $13 \choose x$ variants. We now need to choose without ordering and with replacement one variant for each of this variant, and this number is known as multiset coefficient. Taking formula for it, we get $[x, y] = {{{13 \choose{x}} + y - 1}\choose y}$.
The approach outlined in that post works when all suits have the same number of cards -- as is the case for the first round of dealing. But I'm struggling to translate it to a situation where there are a varying number of cards per suit.
Take a suit configuration to be $(r_0, r_1, \ldots)$ where $r_i$ is the number of cards in a given suit for the $i$th round of dealing. As an example, in text holdem, the first round of dealing is the 2 pocket cards and the second round of dealing is the 3 cards for the flop.
To illustrate the problem, a suit configuration up to the flop in text holdem could be: $(1, 1)(1, 1)(0, 1)$.
Naively, I calculated the number of non-equivalent hands as: $$ \binom{\binom{13}{1}+2-1}{2} * \binom{\binom{13}{1}+3-1}{3} $$
But the first and second suits have only 12 cards left for the flop, while the 3rd suits has the full 13 cards remaining.
What is the best way to account for this?