In the card game Skat, you play with a deck of $32$ cards and each of the three players gets $10$ cards.
This means that there are $32! \approx 2.63 × 10^{35}$ possible permutations of the card deck.
And for each player's hand of $10$ cards, there may be ${{32}\choose{10}} \cdot 10! = 234,102,016,512,000$ possibilities, if this is really a partial permutation.
Is this correct?
But in the related card game Doppelkopf ("double-head"), you play with two decks of $24$ cards each. So you have two queens of clubs, for example, which you cannot distinguish. Each player gets 12 cards.
How do you calculate the numbers in this case? Do we have to divide the number of permutations like this?
$$ \frac{48!}{2!^{24}} \approx 7.39 × 10^{53} $$
And what about the possibilities per each player's hand? I have no idea for that one.
Look at the problem in this way: you have 24 distinct cards and you allow the same card to be drawn twice, but not more. Thus the number ways to pick 12 cards (assuming order doesn't matter, as is usual with card hands) is the number of solutions to the equation $$x_1 + x_2 + ... + x_{24} = 12$$ where each $x_i$ can be $0$, $1$, or $2$. If one considers the expansion of $$ (1 + x + x^2)^{24} $$ one may determine that the coefficient of $x^{12}$ is the number of solutions to the above equation. By applying the binomial theorem twice, one finds $$(1+ x + x^2)^{24} = \sum_{k=0}^{24}\sum_{k=0}^n \binom{24}{n}\binom{n}{k}x^{48-2n+k}$$ $x^{48 - 2n + k}$ is $x^{12}$ when $k = 2n - 36$. Since $k ≥ 0$ and $n ≤ 24$, the only solutions $(n, k)$ to this equation are $(18, 0)$, $(19, 2)$, $(20, 4)$, $(21, 6)$, $(22, 8)$, $(23, 10)$, and $(24, 12)$. Therefore, the coefficient of $x^{12}$ is $$\binom{24}{18}\binom{18}{0} + \binom{24}{19}\binom{19}{2} + \binom{24}{20}\binom{20}{4} + \binom{24}{21}\binom{21}{6} + \binom{24}{22}\binom{22}{8} + \binom{24}{23}\binom{23}{10} + \binom{24}{24}\binom{24}{12}$$
which becomes $$ 134596 + 42504 * 171 + 10626 * 4845 + 2024 * 54264 + 276 * 319770 + 24 * 1144066 + 2704156 = 287134346 \approx 2.87 * 10^8$$
So there are about $2.87 * 10^8$ possible hands for some player playing Doppelkopf.
As for your first question, there are $48!$ permutations, but some of these are indistinguishable. Consider two indistinguishable permutations to be 'equivalent.' In each permutation, there are 24 pairs of duplicates. One may switch any of these pairs and get an equivalent permutation. There are $2^{24} \approx 1.68 * 10^7$ possible ways to choose which pairs to rearrange, so each set of equivalent permutations contains $2^{24}$ total permutations. Thus, the number of distinguishable permutations is as you said $\frac{48!}{2^{24}} \approx 7.40 * 10^{53}$.