Finding an intuitive explanation suppose we have 3 balls and 3 boxes, so we can totally arrange them in 3! (=6) ways. Additionally suppose we have 5 new balls and 5 new boxes and the total number of arrangements will be 5! (=120). Now suppose you put them together. If there is a wall between, the total number gets to 3! x 5! (=720). And the total number of arrangements gets to 8! (=40320), if there will be no wall and all the balls will be allowed to be placed in each of the 8 boxes freely. Well, everything seems to be a no-brainer. But my problem is that I can’t find an intuitive explanation why the wall removal (or allowing all the balls to be places in each of 8 boxes freely) will cause the total number of arrangements (720) to be multiplied by 56 (8x7). Can anybody give an intuitive explanation that I can visually have in my mind? 
Here's a visual explanation of my question: 
Visual explanation of my question!
 A: Your first interpretation is a Cartesian product, and doesn't mean much.
For the second, use stars and bars, theorem 2 to intersperse placeholders for the elements $\{1,2,3\}$ into the permutations of $4,5,6,7,8$.
There are $\binom{8}{3}=8\cdot7=56$ of these, and then $3!=6$ ways of placing $1,2,3$ into the placeholders, so we have $5!\cdot56\cdot6=8!=40320$.
A: Using JMoravitz's hint:
Take an arrangement with the wall. You have an order for the first three balls and for the last five balls. 
Suppose you have eight boxes in a row to put these eight balls (with no wall). Choose three of those boxes and place the first three balls (from the arrangement with the wall) in order from left to right. Take the last five balls and place them in the remaining boxes in order from left to right. This is a distinct permutation without walls.
So, there are $$3!\times 5!\times \dbinom{8}{3} = 8!$$ ways to arrange them.
A: If there is a wall the balls are not allowed into boxes on the other side of the wall.  If there isn't a wall, they are.  That's all.
With the wall, the three balls are only allowed in three boxes and the five balls are only allowed in five boxes.
But without the wall, the three balls are allowed in any of the eight boxes and the five balls are too.
