We can use casework to solve this problem.

First, le'ts consider cases with \(4-0-0\), where all 4 balls are in one box.

Since each box is indistinguiashable, there is only 1 way to complete this.

Second, let's consider cases with \(3-1-0\)

We can calculate this by using \(4 \choose 1 \), so there are 4 ways.

Third, let's consider cases with \(2-2-0\)

There are \(4 \choose 2\) ways to complete this, so there are 6 cases.

Lastly, we have \(2-1-1\)

We have \(4 \choose 2 \) ways, to do this, so 6 additional cases.

Now, we add all of these up. We get \(1 + 4 + 6 + 6 = 17\)

So our final answer is 17.

Thanks! :)