number of arrangements calculation

I have calculated the solution correctly, but my approach is cumbersome and needs to be more general.

I have listed the possible arrangements without distinguishing the objects.

1. (4-0-0, 0-4-0, 0-0-4) => 3 possibilities
2. (3-1-0, 3-0-1, 1-3-0, 0-3-1, 1-0-3, 0-1-3) => 6 possibilities
3. (2-2-0, 2-0-2, 0-0-2) => 3 possibilities
4. (2-1-1, 1-2-1, 1-1-2) => 3 possibilities

I have calculated the possibilities for each arrangement.

1. $3 \times 1$ => there is only one way to place 4 object in one place
2. $6 \times 4$ => there are 4 ways to place 4 distinct objects 3-1-0
3. $3 \times \frac{4 \times 3}{2}$ => etc...
4. $3 \times \frac{4 \times 3}{2}\ \times 2$ => etc...

total = $3 \times 1$ + $6 \times 4$ + $3 \times \frac{4 \times 3}{2}$ + $3 \times \frac{4 \times 3}{2}\ \times 2$ = 81

If anybody can provide a more general solution and explanation, it would be much appreciated.