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.
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
If both objects and boxes are distinguishable:
There are $n$ ways to select a box that'll be empty. Since rest of the boxes will have at least 1 object each, therefore there will one box out of the $n-1$ that will have two objects in it. There are $n-1$ ways to choose this box. Further, we have ${n\choose2}$ ways to put two objects in the selected box and $(n-2)!$ ways to arrange the rest of the objects in the remaining $n-2$ boxes such that each of those boxes gets exactly $1$ object. So the number of ways will be $n(n-1){n\choose2}(n-2)!=n!{n\choose2}$
A) If both objects and boxes are indistinguishable, there will be only $1$ way of placing the objects.
B) If the objects are indistinguishable, there will be only $1$ way of placing the objects after choosing the boxes. So there are $n(n-1)={n\choose2}$ ways.
Another method when both are distinguishable:
We first select $2$ objects that will be together in the box containing $2$ objects. There are $n\choose 2$ ways to do this. We now have $1$ "double object", $1$ "empty object" and $n-2$ normal objects. Since these are obviously distinguishable, we need to place $n$ objects in $n$ boxes. So there are $n!$ ways to do this. Therefore our answer will be $n! {n\choose 2}$. This can be done similarly for (B).