In how many ways can 4 balls be placed in 4 boxes if the balls are indistinguishable, and the boxes are distinguishable?
The balls are all the same so the only differenciating thing is how many balls are in each individual box.
The boxes can have
0,0,0,4 3 possibilities
0,0,1,3 6 possibilities
0,0,2,2 3 possibilities
0,1,1,2 3 possibilities
1,1,1,1 1 possibilities
So that are 3 + 6 + 3 + 3 + 1 = 16 unique outcomes
You got the right idea but got your permutations mixed up.
The boxes can have
0,0,0,4 3 possibilities............4! / 3! ==4
0,0,1,3 6 possibilities............4! / 2! ==12
0,0,2,2 3 possibilities............4! / 2!.2! ==6
0,1,1,2 3 possibilities.............4! / 2! ==12
1,1,1,1 1 possibilities..............4! / 4! ==1
Total ==4 + 12 + 6 + 12 + 1 ==35 ways - which is the same as:
[4 + 4 - 1] C 4 ==7 C 4 ==35 ways
