Permutation Problem: Select m objects from n ones of r types by sampling without replacement

Question

Problem Description

If there are $n$ objects with:

• $n_1$ indistinguishable objects of type 1,
• $n_2$ indistinguishable objects of type 2,
• ...
• $n_r$ indistinguishable objects of type r

s.t. $$n_1 + n_2 + ... + n_r = n$$ How many possible arrangements are there for selecting $m$ objects from them by sampling without replacement, where orders of the same type of objects do not matter while orders of different types matter.

Background

I thought of this question when I was reviewing the counting chapter of the book Discrete Mathematics and Its Applications. When it came to the generalized problems of permutations and combinations, I came up with and got confused of this question which I thought belonged to the box-ball problems. But I could not find a solution in the book, neither in Google. No doubt that this question is interesting and valuable.

Progress

The only idea about the solution is the case of $m=n$, and the solution should be $$\frac{n!}{n_1!n_2!...n_r!}$$

Answer 1

The general open-form solution could be of the form $\sum_{\begin{matrix}k_1,k_2,...,k_r\text{ such that }\\k_1+k_2+...+k_r=m\\k_1\in[0,min\{n_1,m\}]\\k_2\in[0,min\{n_2,m\}]\\...\\k_r\in[0,min\{n_r,m\}]\\k_i integers\end{matrix}}\left(\begin{matrix}m\\k_1k_2...k_r\end{matrix}\right)$.

Writing out each $n_i$ in a grid like format, it remains to find all possible combinations of values that sum to m given differing lengths of $n_1,n_2,...,n_r$, a seemingly challenging task.

$\begin{matrix}0 & 1 & 2 & 3 & \textbf{4} &... & n_1\\0&1&\textbf2&3&4&...&...&n_2\\...\\0&1&2&\textbf{$n_r$}\end{matrix}$

I have found it easy to begin small and consider the case with $\sum_{i=1}^rn_i=m$. In this case, You must pick all the things, and there is only one possibility. Then the answer is the multinomial coefficient with the corresponding values of $k_1=n_1,...,k_r=n_r$.

Now consider when $\sum_{i=1}^rn_i=m+1$. Looking at the hastily drawn image above, there is one degree of freedom that must be distributed across the r rows. There are r ways to do this. Now you must select all r combinations of $k_1,...,k_r$ such that they sum to m, plug that into the multinomial coefficient, and sum them up.

Consider when $\sum_{i=1}^rn_i=m+2$. Depending on r, the answer is different. $\begin{cases}r=1&r\\r\ge2&r+{r\choose2}\end{cases}$. When you've got only one type of item, you must pick all of them. But if you've got at least 2 types of things, you can distribute the 2 degrees of freedom to the same item (r possibilities) or distribute 1 each to 2 items (r choose 2). Note that this does not include actually finding the combinations of the $k_i$, this only tells you how many combinations there are.

Consider $\sum_{i=1}^rn_i=m+3$. Again, there are cases depending on r. $\begin{cases}r=1&r\\r=2&r+2{r\choose2}\\r\ge3&r+2{r\choose2}+{r\choose3}\end{cases}$. Notice the number of cases is how much smaller m is than n.

For $\sum_{i=1}^rn_i=m+4$, $\begin{cases}r=1&r\\r=2&r+2{r\choose2}+{r\choose2}\\r=3&r+2{r\choose2}+{r\choose2}+\left(\begin{matrix}3\\2,1\end{matrix}\right){r\choose3}\\r\ge4&r+2{r\choose2}+{r\choose2}+\left(\begin{matrix}3\\2,1\end{matrix}\right){r\choose3}+{r\choose4}\end{cases}$. Notice that a multinomial coefficient has appeared. I believe there will be at most one multinomial coefficient per "term" if you proceed this way, i.e. you won't have to worry about multinomial coefficients multiplying multinomial coefficients etc.

It has occurred to me that you can also start the other way, with $m=0$ and proceed upward.