Why is the way I solve this wrong: “how many ways can we put $k$ balls in $n$ cells”? So the problem is how many ways can we put $k$ balls in $n$ cells. The correct way my professor solved it was by saying that the first ball can be placed with $n$ different ways and the second with $n$ different ways... and the $k$-th ball with $n$ different ways. But when I think about it, it seems correct to say in the first cell I can place $k$ balls, and in the second one $k$ balls... and in the $n$-th cell I can place $k$ balls. But this gives a different answer: my professor’s give $n^k$ and mine $k^n$. So why is my approach of solving the problem wrong?
 A: There are 2 reasons why the answers differ:
1: Any single cell can contain any number of balls from 0 to k.
2: Every ball must be placed in exactly one cell.
A: So your professor stating that you can place each ball in n different ways is correct because every ball has n places in where it can be placed. I think this is obvious. Your idea of being able to play k balls into 1 cell and k into the next and so on can't work because if you place k balls into the first cell, you won't have any balls to place into the second cell and so on. Now, you are implying that each cell can take up to k balls (which is correct) but this doesn't answer the question of how many ways can k balls be placed into n cells.
If we look at the answer of your professor: $n^k$ implies that you can place one ball into n possible places and you have k number of balls. Hence, $(n)(n)...(n)$ k times $= n^k$
Your method is if we can put k balls into cell 1 and k balls into cell 2 and so on, n times, we would have $(k)(k)...(k) = k^n$. This won't answer your question of how many ways can k balls be split up between n cells. This would be how many ways can you split a group of unique balls into n cells where each cell must hold k distinct balls (essentially partitioning the balls into unique groups). It's the same idea of if you have two pairs of shoes, two pairs of pants and two shirts, how many combinations of these things could you have? The answer being $(2)(2)(2) = 2^3$
