Distinct objects into identical bins with minimum $n$ in each bin The question states that:
9 players are playing a game and they have to gather at 3 identical booths. How many
ways can this be done if each booth must have at least 2 players if the arrangement of the
players does not matter?
Personally, I have tried finding any general formula. I found Stirling number of the second kind but it can only be applied to non-empty groups. In this problem that I am facing, it clearly states that it requires at least 2 players in each booth and hence I am at a loss of how to solve this case, and if possible for my own learning, find a generalised formula.
My solution is that
There can be $3$ ways to arrange $9$ players into $3$ groups with each group having more than $2$ members.
$$\text{Case 1: AA BB CCCCC (2, 2, 5 combination)}$$
$$\text{Case 2: AA BBB CCCC (2, 3, 4 combination)}$$
$$\text{Case 3: AAA BBB CCC (3, 3, 3 combination)}$$
$$\text{Case 1: } \frac {{9 \choose 5} \times {4 \choose 2} \times {2 \choose 2}} {2!} = 378$$
$$\text{Case 2: } {{9 \choose 4} \times {5 \choose 3} \times {2 \choose 2}} = 1260$$
$$\text{Case 3: } \frac { {9 \choose 3} \times {6 \choose 3} \times {3 \times 3}} {3!} = 280$$
Total number of possible ways: $378 + 1260 + 280 = 1918$
I was wondering if this solution is correct. If this solution is correct, is there any generalised formula I could use?
 A: The answer seems to be $1918$, as you found. Here is another way to find it, though there may be a simpler way. Initially order the booths $1$, $2$, and $3$. Then condition on how many people are in booths $1$ and $2$, choose the people going in booths 1,2, and 3, and finally divide by $3!$ since order of booths doesn't matter. This way of thinking leads to the solution
$$\frac{1}{6}\sum_{x=2}^5\sum_{y=2}^{7-x}\frac{9!}{x!y!(9-x-y)!} = 1918.$$
A: This question can be solved by exponential generating functions. You can learn about it in https://www.math.brown.edu/mchan2/2019Spring_2520/StudentPresentations/George_Exponential_Generating_Functions.pdf
We should find the exponential generating function of each booth such that
$$\frac{x^2}{2} + \frac{x^3}{6} +\frac{x^4}{24} + ..$$
Because each booth must be assignated at least two players.
NOTE =
Now , we need to find the expansion of $$\bigg(\frac{x^2}{2!} + \frac{x^3}{3!} +\frac{x^4}{4!} + \frac{x^5}{5!}\bigg)^3$$
After that , find the coefficient of $x^9$ and multiply it by $9!$ or find the coefficient of $\frac{x^9}{9!}$ such that https://www.wolframalpha.com/input/?i=expanded+form+of+%28x%5E2+%2F+2+%2B+x%5E3+%2F+6+%2B+x%5E4+%2F+24+%2Bx%5E5+%2F+120%29%5E3
Then , $$9! \times \frac{137}{4320} = 11508 $$
However , as other experts remind me  , the order of booths does not matter so , i will divide them by $6$ (because the order of booths does not matter such that $ABC=BCA=CBA$ ..etc , there are $3!$ ways to arrange). Then , $$11508 / 6 =1918$$
A: Another approach is to apply Principle of Inclusion Exclusion. Using P.I.E, we count number of ways as,
$\small \displaystyle 3^9 - {3 \choose 1} \left[2^9 + 9 \cdot (2^8 - 2) \right] + {3 \choose 2} \left[1 + 9 \cdot 8 \right] = 11508$
$\small \displaystyle {3 \choose 1} \left[2^9 + 9 \cdot (2^8 - 2) \right]$ is the number of ways for one of the booths to have either no person or one person.
$ \small \displaystyle {3 \choose 2} \left[1 + 9 \cdot 8 \right]$ is the number of ways for two of the booths to have either no person or one person each.
As booths are identical, we divide by $ \small 3!$ and the answer is $\small 1918$.
A: To generalize the other answer, the generating function for $n$ people and $m$ booths, with the restriction that at least $2$ people are in each booth, is $\frac{n!}{m!}$ times the coefficient of $x^n$ in:
$$\begin{align}(e^x-1-x)^m&=\sum_{k=0}^{m}(-1)^{m-k}\binom mke^{kx}\sum_{j=0}^{m-k}\binom{m-k}jx^j\\
&=\sum_{k+j\leq m}(-1)^{m-k}\binom{m}{k,j,m-k-j}e^{kx}x^j\end{align}$$
The coefficient of $x^{n}$ for the $k,j$ term is:
$$\frac{k^{n-j}}{(n-j)!}(-1)^{m-k}\binom m{k,j,m-k-j}$$
Summing and then multiplying by $n!/m!,$ you get:
$$\sum_{k=1}^{m}(-1)^{m-k}k^{n-m}\sum_{j=0}^{m-k} \frac{k^{m-j}}{k!j!(m-k-j)!}(n)_j\tag1$$
where $(n)_j=n(n-1)\cdots(n-j+1)=\frac{n!}{(n-j)!}$ is the “falling factorial.” (We can restrict to $k\neq 0$ since $n>m\geq j$ means $k^{n-j}=0$ when $k=0.$)
For fixed $m,$ the sum here are dominated by:
$$\frac{m^{n}}{m!}$$ for large $n.$
Note, you get a problem with $1/(n-j)!$ in (1) if $j>n,$ but then $m>n,$ and the count should be zero.

When $m=3,$ we have $(k,j)=(1,0),(1,1),(1,2),(2,0),(2,1),(3,0).$
Getting:
$$\left(\frac{1}{2}+n+\frac12n(n-1)\right)-
2^{n-3}\left(\frac{2^3}{2}+\frac{2^2}{2}n\right)+
3^{n-3}\left(\frac{3^3}{6}\right)\\
=\frac{3^{n-1}+n^2+n+1}{2}-2^{n-2}(n+2)
$$
This give $1918$ when $n=9.$

The question gets hard if you require more people in the booth. If you require $3$ people in the booth, then the generating function is:
$$(e^x-1-x-x^2/2)^m$$
which is going to require a lot more terms.

The general function for at least $p+1$ people in each booth can be written:
$$\sum_{k+j_0+\dots+j_{p}=m}(-1)^{m-k}e^{kx}\binom{m}{k,j_0,\dots,j_{p}}\frac{x^{j_1+2j_2+\cdots+pj_p}}{(0!)^{j_0}(1!)^{j_1}\cdots (p!)^{j_{p}}}
$$
If we write $J$ as a shorthand for $j_1+2j_2+\cdots+pj_p$ the result is:
$$\sum_{k=1}^{m}(-1)^{m-k}k^{n}\sum_{j_0+\cdots+j_p=m-k}\frac{1}{k^Jk!\prod_{i=0}^p j_i!(i!)^{j_i}}(n)_J$$
This is thus of the form:
$$\sum_{k}(-1)^{m-k}k^n Q_{m,p,k}(n)$$ where each $Q_{m,p,k}$ is a polynomial of degree $p(m-k).$
Again, asymptotically, theirs is $\frac{m^n}{m!}.$
