Sorry, that is not correct. Thanks for trying.
The correct solution is this:
For the 4, 0, 0 distribution we have only 1 way.
For the 3, 1, 0 distribution we have 4 ways.
The 2, 2, 0 distribtution is where NotThatSmart makes a mistake. It would appear to be \(4 \choose 2\) = 6 ways, but there is double counting. For example, choosing balls 1 and 3 leads to the same split as choosing balls 2 and 4. So, for the 2, 2, 0 distribution there are only 3 ways, making the total amount 14 ways.
