I have a dice with $K$ many outcomes. I am rolling this dice $n$ times. Assume $k_i$ denotes number of times class $i = 1,\ldots,K$ appears in $n$ many rolls. I am wondering, in how many different ways we can assign the $k_i$ values. Here, the order does not matter.
For example, let $n=3$ and $K=2$. We have four different outcomes:
- $k_1 = 3, k_2 = 0$
- $k_1 = 0, k_2 = 3$
- $k_1 = 2, k_1 = 1$
- $k_1 = 1, k_2 = 2$
So in this small example, there are $4$ different ways to assign $k_1,k_2$ values.
I think the problem is equivalent to the number of solutions to the following system: $$k_1 + k_2 + \ldots k_K = n \\ k_i \geq 0, \text{and integer for }, i=1,\ldots,K $$
So the solution is roughly: $$\underbrace{{K}\choose{1}}_{\text{one takes $n$, others $0$}} + \underbrace{{K\choose{1}}}_{\text{one takes $n-1$}}\underbrace{{K-1\choose1}}_{\text{one takes $1$}} + \underbrace{{K\choose{1}}}_{\text{one takes $n-2$}}\left[ \underbrace{K-1\choose 1}_{\text{one takes $2$}} \underbrace{+}_{\text{or}} \underbrace{K-1\choose 2}_{\text{two take $1$}} \right] + \ldots $$ Is there a closed-form answer to this sum?
Your assumption that the problem is equivalent to finding the number of non-negative integer solutions to $$k_1 + k_2 + \ldots k_K = n \\ k_i \in\mathbb{Z}_{\ge0},\; i=1,\ldots,K $$ is correct.
The problem is very well-known and its solution $$ \binom{n+K-1}n $$ can be easily found by stars and bars method.