Take the simplest model: One after the other of the balls is thrown at random into one of the boxes $a$, $b$, $c$. There are $3^{12}$ different possible histories for that, all of them equiprobable. The number of histories leading to a particular final content $(r,s,t)$ of the three boxes is the coefficient of the term $a^r b^s c^t$ in the expansion of $(a+b+c)^{12}$, i.e., is given by ${12!\over r!\>s!\>t!}$.
There is the question whether "one of the boxes" means "at least one of the boxes", or "exactly one of the boxes". Since one sentence later they talk about "exactly three balls" my working hypothesis is that "at least one of the boxes" is meant.
For the probability in question we have to consider the contents
$$(3,9,0), (3,8,1), (3,7,2), (3,5,4)$$
each of them in six orders, and $(6,3,3)$ in three orders. The total number $N$ of "admissible" histories is therefore given by
$$N=6{12!\over3!}\left({1\over9!}+{1\over8!}+{1\over 7!\>2!}+{1\over 5!\>4!}\right)+3{12!\over 6!\>3!\>3!}=282\,480\ ,$$
and the required probability $P$ is $$P={N\over 3^{12}}\doteq0.531536\ .$$
There are three choices for each of the five balls. Hence, if there were no restrictions, the balls could be placed in the boxes in $3^5$ ways. From these, we must exclude those distributions in which one or more of the boxes is empty.
There are $\binom{3}{1}$ ways to exclude one of the boxes and $2^5$ ways to distribute the balls to the remaining boxes. Hence, there are
$$\binom{3}{1}2^5$$
ways to distribute the balls so that one of the boxes is empty.
However, we have counted those distributions in which two of the boxes are empty twice, once for each of the ways we could have designated one of the empty boxes as the excluded box. We only want to exclude them once, so we must add these cases back.
There are $\binom{3}{2}$ ways to exclude two of the boxes and one way to place all the balls in the remaining box.
Hence, the number of ways the balls can be distributed so that no box is left empty is
$$3^5 - \binom{3}{1}2^5 + \binom{3}{2}1^5$$
by the Inclusion-Exclusion Principle.
Where am I going wrong?
You count each distribution in which one box receives three balls and the others receive one three times, once for each way you could place one of those three balls first.
You count each distribution in which two of the boxes receive two balls and the other box receives one four times, once for each way you could place one of the two balls in each of the two boxes with two balls first.
Three balls in one box and one ball in each of the others: There are three ways to choose which box receives three balls, $\binom{5}{3}$ ways to choose which three balls are placed in that box, and $2!$ ways to distribute the remaining balls. Hence, there are
$$\binom{3}{1}\binom{5}{3}2!$$
ways to distribute the balls so that three balls are placed in the same box.
Two boxes receives two balls and one box receives one ball: There are three ways to choose which box receives only one ball and five ways to choose the ball that is placed in that box. There are $\binom{4}{2}$ ways to choose which two of the remaining four balls are placed in the smaller of the two remaining boxes. The other two balls must be placed in the remaining box. Hence, there are
$$\binom{3}{1}\binom{5}{1}\binom{4}{2}$$
ways to distribute the balls so that two boxes receive two balls and one box receives one.
Observe that
$$\binom{3}{1}\binom{5}{3}2! + \binom{3}{1}\binom{5}{1}\binom{4}{2} = 3^5 - \binom{3}{1}2^5 + \binom{3}{2}1^5$$
Since you counted distributions in which one box receives three balls and the others receive one three times and distributions in which two boxes receive two balls and the other receives one four times, you obtained
$$3\binom{3}{1}\binom{5}{3}2! + 4\binom{3}{1}\binom{5}{1}\binom{4}{2} = \binom{5}{3} \cdot 3! \cdot 3^2$$
Best Answer
When we work over probability it does not matter whether the balls are distinct or identical. Beucause , when we select one ball to place any box , the selected ball become distict , because it is the first ball selected. Did you get the idea ? It is not anymore identical , because it is in our hands to place and it is the first one.It is valid for the rest , for example the second selected ball , it was the same as the others ,but when we take it from among the others , it become suddenly distinguishable , it become distinguishable because , it is "the second selected ball",now we are able to distinguish it !. Now , you have all distinct balls when you place them , because all of have selection order to be distinguished.
When we comes to your question , we have $12$ balls each have different selected order ,i.e they can be distinguished from other by their selection order. Select $3$ of them to satify given condition. These balls can be second selected ball , sixth selected and tenth selected or third selected , fifth selected and seventh selected. So , you just need to select $3$ order among $12$ order by $C(12,3)$