I have 4 decks, each with five unique cards. If I select 3 from the first, 4 from the second, 2 from the third and 2 from the fourth. In case order matters and I do not put cards back, in how many ways can I arrange the 11 cards if the order matters for each selected card?
I think that I use the following formula, for each deck, and just multiply them.
$\displaystyle\frac{n!}{(n-k)!}$
I think it should be -
$4! × \binom 53 × \binom 54 × \binom 52 × \binom 52$
I take it that the $11$ cards selected can be in all possible orders,
not necessarily with the cards of each deck together, then
Number of ways $=\dbinom53\dbinom54\dbinom52\dbinom52 \times \dfrac{11!}{3!4!2!2!}$
If order matters then instead of binomial coefficients we must use just falling factorials. In combinatorics these kind of counts are called variations without repetitions, there are a special kind of permutations, that is, choosing $k$ different objects from $n$ different objects then all possible orderings are
$$n^\underline k:=\prod_{j=0}^{k-1}(n-j)=n\cdot (n-1)\cdots (n-k+1)$$
hence from the first deck we have $n_1^\underline{k_1}$ possible outcomes, from the second $n_2^\underline{k_2}$, etc... (in your example $n_1=n_2=\ldots=5$, and $k_1=3$, $k_2=4$ and so on).
Then if you choose from your $m$ different decks with a prefixed order in the decks (as it seems in your example) then all possible orderings are
$$A:=\prod_{j=1}^m n_j^\underline{k_j}=n_1^\underline{k_1}\cdot n_2^\underline{k_2}\cdots n_m^\underline{k_m}$$
If you choose the order of the decks randomly then all possible orderings are $A\cdot m!$.