$8$ distinct balls are randomly distributed among $4$ boxes. What is the probability that each box has exactly two balls? I made up an question in my head as to probability but i am not sure about the solution.
Question: There are $8$ distict balls and $4$ distinct boxes.An individual distributes the balls into boxes randomly.What is the probability that each boxes have exactly two balls?
My solution: The sample space is ${4}^{8}$   and the number of ways distributing exactly two balls into each four boxes : $ C(8,2).C(6,2).C(4,2).C(2,2).P(4,4)$
$\therefore \frac {C(8,2).C(6,2).C(4,2).C(2,2).P(4,4)}{{4}^{8}} $
Is my solution correct? I feel that i am wrong in somewhere.If it is not correct ,can you give hints or the solution. Thanks for your helps..

A natural interpretation is that you have a sequence of eight steps: at step $n$ you take ball number $n$ and put it in some numbered box.
You can then choose two steps among the eight at which a ball will be placed in box number $1$. There are $C(8,2)$ ways to choose the two steps, which is tantamount to choosing which two balls go in box $1$.
Next you choose two steps among the remaining six at which a ball will be placed in box number $2$. There are $C(6.2)$ ways to choose these two steps.
Next you have have $C(4,2)$ ways to choose the two steps at which to put a ball in box $3$, and finally $C(2,2) = 1$ way to choose the two steps at which to put a ball in box $4$.
Notice that among the $C(8,2) C(6,2)$ ways the balls could go into boxes $1$ and $2$, one way is to put balls $1$ and $2$ in box $1$ and to put balls $3$ and $4$ into box $2$.
Another way is to put balls $3$ and $4$ into box $1$ and to put balls $1$ and $2$ in box $2$.
That is, by the time you have multiplied $C(8,2) C(6,2)$, you not only have counted the number of ways to choose two pairs of balls from the original eight balls, you have also counted the number of ways those two pairs could be placed in the two boxes by switching which pair goes in which box.
You do not have to multiply by any additional factors to account for the ways to distribute the pairs of balls between the two boxes after selecting the pairs.
In fact, if you do multiply by any additional factor other than $1$, you will produce a wrong answer.
Similarly, by the time you have multiplied $C(8,2) C(6,2) C(4,2) C(2,2)$,
you have already counted all the ways to assign the pairs of balls to specific boxes, for example, balls $1$ and $2$ in box $1$, balls $3$ and $4$ in box $2$,
balls $5$ and $6$ in box $3$, and balls $7$ and $8$ in box $4$.
Given any set of four disjoint pairs of balls, you have already counted each permutation of the pairs.
If you now start shuffling pairs of balls among the boxes, you will be multiply counting outcomes that you have already counted.
