I think I have an ugly solution.
instead of all integers being greater than 1, let's say they are all greater than 0. This does not affect the problem at all.
We can define the function f(n,b), the combinations of all the numbers, implicitly with
f(n,b) = f(n-1,b) + f(n-1,b-1) + f(n-1,b-2) ... f(n-1,2) + f(n-1,1) + f(n-1,0)
We want an explicit solution, so we will continue, until everything is of the form f(0,x), since f(0,x) = 1.
f(n,b) = f(n-2,b) + 2*f(n-2,b-1) + 3*f(n-2,b-2) ... (b-1)*f(n-2,2) + b*f(n-2,1) + (b+1)*f(n-2,0)
so the general case is:
f(n,b) = ((i-1)C(0)f(n-i,b) + ((i)C(1))*f(n-i,b-1) +((i+1)C(2))*f(n-i,b-2) ... ((b+i-3)C(b-2))*f(n-i,2) + ((b+i-2)C(b-1))*f(n-i,1) + ((b+i-1)C(b)*f(n-i,0)
so we now extend this until i=n, and we have out explicit formula:
f(n,b) = ((n-1)C(0))*(1) + ((n)C(1))*(1) +((n+1)C(2))*(1) ... ((b+n-3)C(b-2))*(1) + ((b+n-2)C(b-1))*(1) + ((b+n-1)C(b))*(1)
cleaned up a bit:
f(n,b) = (n-1)C(0) + (n)C(1) +(n+1)C(2) ... (b+n-3)C(b-2) + (b+n-2)C(b-1) + (b+n-1)C(b)
This can probably be simplified even further, and I've probably made a fatal mistake somewhere. (it's not f(0,x)=0, I'm pretty sure of that one, even though you can have a hard time rearranging 0 elements)
