There are 5 essentially different ways to distribute the black balls. In each case I'll count the essentially different ways of distributing the white balls.
4, 0, 0, 0
The possible distinguishable ways of distributing the white balls are: 2-0-0-0, 1-1-0-0, 0-2-0-0 and 0-1-1-0. So 4 is the number.
3, 1, 0, 0
Here we can do it like this: 2-0-0-0, 1-1-0-0, 1-0-1-0, 0-2-0-0, 0-1-1-0, 0-0-2-0, 0-0-1-1. So 7.
2, 2, 0, 0
Once again: 2-0-0-0, 1-1-0-0, 1-0-1-0, 0-0-2-0, 0-0-1-1. 5 ways.
2, 1, 1, 0
And again: 2-0-0-0, 1-1-0-0, 1-0-0-1, 0-2-0-0, 0-1-1-0, 0-1-0-1, 0-0-0-2. 7 ways.
1, 1, 1, 1
Lastly: 2-0-0-0, 1-1-0-0. 2 ways.
In total, $4+7+5+7+2 = 25$ ways to distribute the balls.
If we let s be the number of ways to do this, then the number of ways to distribute 6 distinguishable objects into 4 distinguishable boxes so that no box is empty is given by $4!(s)=24s$,
$\;\;\;$since there are $4!$ ways to label the boxes.
We can also count the number of ways to distribute the objects into 4 distinguishable boxes so that no box is empty using Inclusion-Exclusion:
If $T$ is the set of all distributions, and $A_i$ is the set of distributions with box i empty, for $1\le i\le 4$,
then $\lvert A_1^c\cap\cdots\cap A_4^c\lvert=\lvert T\lvert-\dbinom{4}{1}\lvert A_1\lvert+\dbinom{4}{2}\lvert A_1\cap A_2\lvert-\dbinom{4}{3}\lvert A_1\cap A_2\cap A_3\lvert$
$\hspace{1.3 in}=\displaystyle 4^6-4\cdot3^6+6\cdot2^6-4\cdot1^6=1560$,
so $24s=1560$ and therefore $s=65$.
This answer is $S(6,4)$, a Stirling number of the second kind.
Best Answer
OK, so I found a useful resource that helped me understand (I think) how to apply the idea of partitioning to this problem.
Before starting, I should say that we will assume each box can contain all 8 objects, no objects, or any integer in between. Other answers may use other assumptions.
Basically:
Definition: P(k, i) is the number of partitions of k into i parts.
In this case P(8,i), where i $\leq 8$.
P(k) is the number of partitions of k into any number of parts.
In this case we want P(8).
So we need to enumerate all the possible combinations:
P(8,1) = 1 (i.e. 8 objects in one box, all others empty.)
P(8,2) = 4 [(6,2), (7,1), (5,3), (4,4)]
P(8,3) = 5 [(5,2,1), (3,3,2), (4,2,2), (4,3,1), (6,1,1)]
P(8,4) = 5 [(2,2,2,2), (4,2,1,1), (3,2,2,1), (5,1,1,1), (3,3,1,1)]
P(8,5) = 3 [(4,1,1,1,1), (3,2,1,1,1), (2,2,2,1,1)]
P(8,6) = 2 [(3,1,1,1,1,1), (2,2,1,1,1,1)]
P(8,7) = 1 [(2,1,1,1,1,1,1)]
P(8,8) = 1 [(1,1,1,1,1,1,1,1)]
P(8) = $\sum{P(8,i)}$ = 22
Therefore the answer is 22.
EDIT: "A recurrence relation is part(n,k) = part(n-1,k-1) + part(n-k,k)" This is from: http://2000clicks.com/mathhelp/CountingObjectsInBoxes.aspx
The site contains a link to the proof.