7 different balls put into 4 identical boxes, how to count? 7 balls with different colors, there are 4 identical boxes. the box can be empty. how many ways to distribute the balls?
what kind of counting problem is this? how do we count it?
 A: We have $7$ objects (let each be elements of a set) and we want to partition these objects into $4$ subsets. We can have $1$ nonempty subset (and the other $3$ consequently empty), $2$ nonempty subsets (and the rest empty), $3$ nonempty subsets (and the rest empty), or $4$ nonempty subsets. We can't have $0$ nonempty subsets because that would mean we have no balls in all of the boxes.
The striling number of the second kind $S(n,k)$ counts the number of ways to partition a set of $n$ objects into $k$ non-empty subsets. So the answer is as you claim,
$$S(7,1)+S(7,2)+S(7,3)+S(7,4)$$
A: From set A (the m balls) to set B (the n boxes), the number of onto functions is $n!S(m,n)$. If the boxes are identical, the order does not matter anymore. Therefore, there are $n!S(m,n)/n! = S(m,n) $ ways to distribute the 7 balls. 
We can use, 4 boxes, 3 boxes, 2 boxes or 1 box. 
$S(7,4)+S(7,3)+S(7,2)+S(7,1)$
A: The answer can be obtained w/o using Stirling numbers as
$1+63+\dfrac{(4^7 -1*4 - 63*12)}{4!}= 715$, as explained below
If all configurations had $4!$ permutations, we'd get $\;\dfrac{4^7}{4!}$, but two types don't, and we are adjusting for them

*

*Just $1$ way to place all balls in one box, $4$ ways only to permute, so we add $1$ and subtract $1*4$ in the numerator


*$\binom71 +\binom72+\binom73 = 63$ ways to place balls in two boxes,$12$ ways to permute, so we again add an adjust similarly


*Note that all boxes filled will, of course, have $4!$ permutations, but so will $3$ boxes filled and one blank, because the blank also assumes a distinct identity
