Permutation and combination

Using Separator MEthod:

Balls = 7
Separators = 4

(7+4)!/7!*4!

= 330

The problem as posted first describes the balls as different but then describes them as identical.
I assume that the intention of the problem is as follows:

There are 5 different boxes and 7 identical balls. All of the balls are to be distributed among the boxes that any box can receive any number of balls. In how many ways can the balls be distributed among the boxes?

a.110
b.220
c.330
d.440

OA:C

The following is called the SEPARATOR method.

Since there are 7 identical balls and 5 boxes, the 7 balls are to be separated into -- at most -- 5 groupings.
Thus, we need 7 balls and 4 separators:
O|O|O|O|OOO

Each arrangement of the elements above represents one way to distribute the 7 balls among 5 boxes A, B, C, D and E:
O|O|O|O|OOO = A gets 1 ball, B gets 1 ball, C gets 1 ball, D gets 1 ball, E gets 3 balls.
OOOOOO||||OO = A gets 5 balls, E gets 2 balls.
OOOOOOOO|||| = A gets all 7 balls.
And so on.

To count all of the possible distributions, we simply need to count the number of ways to arrange the 11 elements above (the 7 identical balls and the 4 identical separators).
The number of ways to arrange 11 elements = 11!.
But when an arrangement includes identical elements, we must divide by the number of ways to arrange the identical elements.
The reason:
When the identical elements swap positions, the arrangement doesn't change, reducing the total number of unique arrangements.
Thus, we must divide by 7! (the number of ways to arrange the 7 identical balls) and by 4! (the number of ways to arrange the 4 identical separators):
11!/(7!4!) = 330.

The correct answer is C.

For two similar problems, check here:

https://www.beatthegmat.com/algebra-t215423.html#622071