Q:

How to find effective partition of $n$ into $k$?

Here is a type of question that I find quite often on MO sites, that I couldn't quite solve:
How many ways can I put $n$ identical balls into $k$ identical boxes, with $n>>k$, such that each box has at least one ball?
For me, this question had $n=600$, and $k = 3$. At first, I thought this was a combinations problem, as we could have 3 balls put in each box first, and then the other $600-3=597$ balls can be placed in any boxes. The answer I first thought of was $3^{597}$. I then realised this was wrong, because I was considering each as a separate ball, in which case it wasn't.
Afterwards, I did a little research, and I thought this question might have something to do with compositions of numbers. However, the equation provided, which is $\binom{n-1}{k-1}$ was not the correct answer, when I plugged in $n$ and $k$ respectively, which got me $179101$.   I realised that in my question, the $k$ boxes were not distinguishable, however in the equation, the $k$ boxes are. 
I believe the way I need to solve this problem is using partitions, instead of compositions, of the numbers, but I am not yet sure. I don't know how I can solve this problem with partitions of numbers.
P.S the answer is $30,000$

A:

Note that the $\binom{599}{2}$ compositions are almost exactly $6$ times the correct answer. This makes sense, because in general each partition into three parts corresponds to $3! = 6$ compositions. However, the ones with repeated parts correspond to fewer compositions, so we can't just divide by $6$.
There are three cases for a composition $\lambda_1 + \lambda_2 + \lambda_3$:

All different (six compositions per partition)
Two the same (three compositions per partition)
All the same (one composition per partition)

The sum of all three cases comes to the aforementioned $\binom{599}{2}$.
The third case is a special case of the second. The total ($600$) minus the odd one out must be even, and can be any even number from $2$ to $598$, so we have $299$ partitions, of which one is the third case.
This gives a final count of $$\frac16 \left(179101 - 298 \times 3 - 1 \right) + 299 = 30000$$ as expected.
Obviously this is messier when $k > 3$.

