Permutation of $h$ cards drawn from a deck of $n$ cards made up of $r$ types $h$ cards are drawn from a deck of $n$ cards, made up of $r$ types: $n_1$ cards of the $1^{st}$ type, $n_2$ cards of the $2^{nd}$ type, ..., $n_r$ cards of the $r^{th}$ type.
Reasoning,

*

*if $h = 1$ then the possible hands (permutations of $h=1$ card drawn from $n$ cards) are: $r$ ;


*if $h = n$ then the possible hands (permutations of $h=n$ card drawn from $n$ cards) are: $\frac{n!}{n_1!n_2!n_3!\cdots n_r!}= \prod_{i=1}^{r} \frac{n!}{n_i!}$ ;


*if $r=n$ (that means $n_1=n_2=...=n_r=1$) then the possible hands (permutations of $h$ card drawn from $n=r$ cards) are: $\frac{n!}{(n-h)!}$ .
(Post Scriptum: always taking into account duplicates)
How do you calculate the permutations of $h$ cards drawn from n cards (consisting of r types, where the type $i^{th}$ is present in $n_i$ copies) when $1 ≤h ≤n$? The answer, of course, must be correct for the extreme cases where $h=1$ and $h=n$.
I would like an answer with proof. If there is an answer that requires the use of generative functions, $\color{red}{\text{I ask you to explain it step by step}}$.
Reasoning a bit I think the answer could be similar to  $\frac{n!}{(n-h)!} \frac{1}{n_1!n_2!n_3!\cdots n_r!}= \prod_{i=1}^{r} \frac{n!}{(n-h)!} \frac{1}{n_i!}$ , beasuse it converges to the last two cases discussed if $h=n$ of if $r=n$, but is incorrect for the first case.
 A: One way to get a (not "simple") general expression is the following.
Let $r_k$ be the number of types of cards such that their number is greater or equal than $k$, i.e.:
$$r_k = |\{j: 1 \le j \le r \land n_j \ge k\}|$$
We have that $r_1 = r$.
Given $h$, we can compute the number of wanted permutations for one solution of the equation:
$$\sum_{k=1}^{n}{ki_k}=h \tag{1}\label{1}$$
where $i_k$ is the number of distinct types of cards appearing each exactly $k$ times in the draw.
Starting from $k=n$, there are ${r_n \choose i_n}$ ways of choosing the $i_n$ types of cards. Then, for $k=n-1$ there are ${r_{n-1}-i_n \choose i_{n-1}}$ ways of choosing the $i_{n-1}$ types of cards, and in general ${r_k-\sum_{m=k+1}^{n}{i_m} \choose i_k}$ ways of choosing the $i_k$ types of cards.
Starting from $k=1$, the number of ways to choose $i_1$ distinct cards among the $h$ drawn is:
$${h \choose 1}{h-1 \choose 1}\ldots{h-i_1+1 \choose 1}=\frac{h!}{(h-i_1)!} \tag{2}$$
For $k=2$ we are left with $h-i_1$ cards for choosing $i_2$ couples of distinct types in the following number of ways:
$${h-i_1 \choose 2}{h-i_1-2 \choose 2}{h-i_1-4 \choose 2}\ldots{h-i_1-2(i_2-1) \choose 2} \tag{3}$$
and so on, giving the following total:
$$\prod_{k=1}^{n}\biggl[{r_k-\sum_{m=k+1}^{n}{i_m} \choose i_k}\prod_{j=0}^{i_k-1}{{h-\sum_{m=1}^{k-1}{mi_m}-kj} \choose k}\biggr] \tag{4}$$
where we can express a factor as a multinomial coefficient:
$$\prod_{k=1}^{n}\prod_{j=0}^{i_k-1}{{h-\sum_{m=1}^{k-1}{mi_m}-kj} \choose k}={h \choose h_1,h_2,\ldots,h_s} \tag{5}$$
where $s=\sum_{k=1}^{n}{i_k}$ and $h_1=h_2=\ldots=h_{i_1}=1$, $h_{i_1+1}=h_{i_1+2}=\ldots=h_{i_1+i_2}=2$, $h_{i_1+i_2+1}=h_{i_1+i_2+2}=\ldots=h_{i_1+i_2+i_3}=3$, and so on.
And summing over all solutions of \eqref{1} we get the final result:
$$\sum_{\sum_{k=1}^{n}{ki_k}=h}\;\prod_{k=1}^{n}\biggl[{r_k-\sum_{m=k+1}^{n}{i_m} \choose i_k}\prod_{j=0}^{i_k-1}{{h-\sum_{m=1}^{k-1}{mi_m}-kj} \choose k}\biggr] \tag{6}$$
Note that we can sum over all solutions because when $0 \le r_k-\sum_{m=k+1}^{n}{i_m} \lt i_k$ the binomial coefficient ${r_k-\sum_{m=k+1}^{n}{i_m} \choose i_k}$ evaluates to $0$. If instead $r_k-\sum_{m=k+1}^{n}{i_m} \lt 0$ then, since $r_{k+1} \le r_k$, we will have $r_{k+1}-\sum_{m=k+2}^{n}{i_m}-i_{k+1} \le r_k-\sum_{m=k+1}^{n}{i_m} \lt 0$ and therefore $r_{k+1}-\sum_{m=k+2}^{n}{i_m} \lt i_{k+1}$. Now, if $0 \le r_{k+1}-\sum_{m=k+2}^{n}{i_m}$ we are done, because the corresponding binomial coefficient evaluates to $0$, otherwise we continue to go up till we have $0 \le r_n \lt i_n$ and ${r_n \choose i_n} = 0$.
When $h=1$ the multinomial coefficient simplifies to $1$, we have the only solution $i_1=1$, $i_2=i_3=\ldots=i_n=0$ to \eqref{1}, and therefore the solution is $r_1=r$ as expected.
When $r=n$ we have $r_1=n$, the only solution $i_1=h$, $i_2=i_3=\ldots=i_n=0$ to \eqref{1}, and the multinomial coefficient evaluates to $h!$, therefore the solution is ${n \choose h}h!=\frac{n!}{(n-h)!}$ as expected.
