There are 5 cases:
4 - 0 - 0 - 0
3 - 1 - 0 - 0
2 - 2 - 0 - 0
2 - 1 - 1 - 0
1 - 1 - 1 - 1
For the first case, there is only 1 way.
In the second case, there are \({4 \choose 3} = 4 \) ways.
In the third case, there are \({4 \choose 2} = 6\) ways.
In the fourth case, there are \({4 \choose 2} \times 2 = 12\) ways.
For the fifth case, there is only 1 way.
So, there are \(1 + 4 + 6 + 12 + 1 = \color{brown}\boxed{24}\) ways.
