How many ways can you distribute 4 identical balls among 4 different boxes?
+92 Ok, so here is my version of the answer to this problem.
We can determine that we can split the balls into the groups 4-0-0-0, 3-1-0-0, 2-2-0-0, 2-1-1-0, and 1-1-1-1. Now we have to count how many ways the boxes can be chosen.
For 4-0-0-0, there are 4 ways we can choose the box that gets all four balls.
For 3-1-0-0, there are 4 ways we can choose the box that gets three balls, then 3 ways we can choose the box that gets one ball.
For 2-2-0-0, there are \(\binom{4}{2} = 6\) ways we can choose the two boxes that get two balls each.
For 2-1-1-0, there are 4 ways we can choose the box that gets two balls, then 3 ways we can choose the box that gets no balls (then the other two boxes get one ball each).
For 1-1-1-1, there is only one way.
The total number of ways is then \(4 + 4 \cdot 3 + 6 + 4 \cdot 3 + 1 = \boxed{35}.\)
Hope this helped! 
