I have 9 marbles, which are randomly put, one by one, in 6 different boxes. Each box has equally chance of receive the marble. A box can be empty at the end, but will never contains more than 3 marbles. If a box with 3 marbles is chosen, then another box is selected, repeated until a box with less than 3 marbles is selected.
My question is, how can I determine every solution, and the different combinations for each solution. I found these results, but I really am not sure of what I found:
3-3-3-0-0-0 : ${6 \choose 3} = 20$ combinations
3-3-2-1-0-0 : ${6 \choose 2} \cdot {4 \choose 1} \cdot {3 \choose 1} = 180$
3-3-1-1-1-0 : ${6 \choose 2} \cdot {4 \choose 3} = 60$
3-2-2-2-0-0 : ${6 \choose 1} \cdot {5 \choose 3} = 60$
3-2-2-1-1-0 : ${6 \choose 1} \cdot {5 \choose 2} \cdot {3 \choose 2} = 180$
3-2-1-1-1-1 : ${6 \choose 1} \cdot {5 \choose 1} = 30$
2-2-2-2-1-0 : ${6 \choose 4} \cdot {2 \choose 1} = 30$
2-2-2-1-1-1 : ${6 \choose 3} = 20$
_t_o_t_a_l_ : $20 + 180 + 60 + 60 + 180 + 30 + 30 + 20 = 580$ combinations
Following question: assuming that previous results are corrects, is it true to say this?
There is $\frac{20}{580}$ chance to get 3 boxes empty at the end.
There is $\frac{530}{580}$ chance to have at least one box with 3 marbles at the end.