Imagine you are choosing 2 cards each from different 2 different suits (13 cards in each) and the ordering of suits does not matter. The goal is to calculate the number of possible combinations of cards. The "absolute value" of the suit doesn't matter, e.g. 9♣10♣J♠Q♠ is the same as 9♠10♠J♣Q♣. But neither are the same as 9♠J♠10♣Q♣ (the 9 and 10 are no longer the same suit).
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. Using the multiset coefficient, we get ${{{13 \choose{x}} + 2 – 1}\choose 2} = 3081$ (details).
I would like to arrive at the same answer using a generating function. Naively, I'd expect the coefficient for $x^4$ of $(1+x+x^2)^{13}$ to be the number of combinations — we're constraining each of the 13 ranks to be chosen 0-2 times and choosing 4 cards. But this evaluates to 1651.
What causes the difference? How could the generating function be changed to be correct?
The goal of expressing this using a generating function is to find a solution for the case where there are a differing number of cards per suit.
Best Answer
Your generating function undercounts the combinations because it does not account for the need to choose these cards from two different suits. It includes the combinations where all four cards could be from the same suit or from more than two suits which is not allowed in your problem statement.
So ideally to choose 2 cards from each of two suits the combined generating function would be $((1+x+x^2)^{13})^2$. The coefficient of $x^4$ is 23075 there. But since this counts each combination twice, once for each suit, we'd normally divide by 2. Which gives us 11537.5 which is meaningless. So unfortunately using a generating function here is not possible as far as I can tell.
Using the multiset coefficient method seems like the way to go.
When to use generating functions:
When not to use generating functions: