Distributing $n$ distinct objects into $m$ types of urns with $k_1,k_2...k_m$ urns of each type

Question

I came accross this (rather complex?) combinatorial problem:

I have $18$ distinct objects, $3$ red urns, $7$ blue urns, and $11$ green urns. In how many ways can I distribute the objects into those urns?

The urns of one colour are identical to each-other. I haven't actually solved it yet, and yet I have decided to generalize:

I have $n$ distinct objects, $k_1,k_2,k_3,...,k_m$ urns of types $1,2,3,...,m$. In how many ways can I distribute the objects into those urns?

I think the solution is strongly related to both partitions of $n$ into $m$ parts where order matters, and the stirling numbers of the second kind. But whichever way I take it somewhat reaches a dead-end, or becomes very complex. How do I approach this?

Answer 1

You can do this in the following number of ways $$\sum _{n_1+\cdots +n_m = n}\binom{n}{n_1,n_2,\cdots ,n_m}\prod _{i=1}^{m}\left (\sum _{j = 0}^{k_i}{n_i\brace j}\right ),$$ which corresponds to selecting each set that you are gonna put in the color urn i.e., the $i$-th color gets $n_i$ objects, and then you need to partition them inside the urns(which are indistinguishable).

Edit: One first divides the $n$ objects into $m$ colors. To do that, you select how many of each color are you taking from the pool of $m$ colors, that is, you are considering a composition of $n$ of size $m$ i.e., $(n_1,\cdots ,n_m)$ such that $n_1+\cdots +n_m=n$. Now, when you have colored your objects for each color you have to split them into the indistinguishable urns, here is where you use Stirling numbers, because you cannot distinguish in between the urns and so the only thing that matters is how these objects are distributed in those $k_i$ urns. They could have all been in $1$ non-empty urn or $2$ non-empty urns or...or $k_i$ non-empty urns and that is why you are adding over Stirling numbers. Multiplication gives all possibilities across all colors and summation over compositions gives you the result. Here you have a code(just change the parameters in the last line) to test in sage for your example it gives $4939393366735084$:

s=0
n=0
m=0
k = []
nn = []
def tal(a,c):
    global k,nn,s,n,m
    if a==m:
        if c==n:
            ss = factorial(n)
            for i in range(0,m):
                ss/=factorial(nn[i])
                tt = 0
                for j in range(0,k[i]+1):
                    tt+=stirling_number2(nn[i],j)
                ss*=tt
            s+=ss
            #print(nn,ss)
        return
    for i in range(0,n+1):
        if c+i>n:
            break
        nn[a]=i
        tal(a+1,c+i)
def dale(nnn,mm,kk):
    global nn,n,m,k,s
    n = nnn
    m = mm
    k = kk
    nn = [0 for i in range(m)]
    s = 0
    tal(0,0)
    return s
print(dale(18,3,[3,7,11]))