Lets assume both children and balloons are distinguisable, labeled. Then the number of distributions corresponds to selecting a 20 digit sequence of numbers 1 to 6, giving $6^{20}$ possibilities. Let $E_k$ be the event that child k does not receive a ballon, this event corresponds to selecting a 20 digit sequence not containing the number k giving $5^{20}$ possibilities.
$$P(\cup_k E_k) = \sum_k P(E_k) - \sum_{k,l} P(E_k \cap E_l) + \sum_{k,l,m} P(E_k \cap E_l \cap E_m) \dots$$
$$ P(\cup_k E_k) = \sum_{n=1}^5 (-1)^{n+1}\frac{\left(\begin{matrix} 6 \\ n \end{matrix}\right)(6-n)^{20}}{6^{20}} = $$
$$ 6 \left(\frac{5}{6} \right)^{20} - 15 \left(\frac{4}{6}\right)^{20} + 20 \left(\frac{3}{6} \right)^{20} - 15 \left( \frac {2}{6} \right) ^{20} + 6 \left( \frac{1}{6} \right)^{20} $$
First choose the boys who get nothing: $10\choose 2$.
Then, you have to distribute $10$ toys to $8$ boys, in such a way that each boy gets at least one. There will be either one boy with $3$ toys, either two boys with $2$ toys.
In the first case, choose the toys ($10\choose3$) and the boy ($8\choose1$), then distribute the $7$ remaining toys to the $7$ boys: $7!$.
In the second case, chose the first pair of toys and a boy (${10\choose2}{8\choose1}$) then the second pair and the second boy (${8\choose 2}{7\choose 1}$), but since you can exchange them, you count cases twice, so divide by $2$: $\frac12{10\choose2}{8\choose1}{8\choose2}{7\choose1}={10\choose2}{8\choose2}{8\choose2}$. Then there remains the $6$ toys given to $6$ boys, so $6!$ cases.
All in all, the number of cases is:
$${10\choose2}\left[{10\choose3}{8\choose1}7!+{10\choose2}{8\choose2}{8\choose2}6!\right]=1\textrm,360\textrm,800\textrm,000$$
To answer the comment: in the second case, you choose a first pair $\{A,B\}$ of toys, and a boy $X$ who will get it. Then you choose a second pair $\{C,D\}$ of toys among the remaining toys, and another boy $Y$ among the remaining $7$. But, among all such choices, there is one where the pair $\{C,D\}$ is chosen first, for the boy $Y$, and then the pair $\{A,B\}$ is chosen, for the boy $X$. Thus all choices appear twice in the possible outcomes, one in some order, one by exchanging them.
More generally, imagine that $k$ boys will get $p$ toys each. Then for all possible $(b_1,t_1),\dots,(b_k,t_k)$ (here $b_i$ denotes the boy, and $t_i$ the set of toys), any permutation of them also appears in the possible outcomes, though it's really the same combination of gifts. Hence you count them $k!$ each.
Best Answer
Please note that $ \displaystyle {12 \choose 4} {8 \choose 4} {4 \choose 4}$ already orders the groups. So you should not multiply by $3!$ again. Using multinomial theorem, you can directly write it as
$ \displaystyle \frac{12!}{4! \cdot 4! \cdot 4!}$
Say you are distributing toys to Jenny, Mike and John and start by selecting $4$ toys out of $12$ to give to Jenny, $4$ out of remaining $8$ to Mike and finally John gets what is left. In one of the possible distributions, you first select $T_1 - T_4$ for Jenny and from the remaining toys, $T_5 - T_8$ for Mike but you will have another distribution where as part of ${12 \choose 4}$, you choose $T_5 - T_8$ for Jenny and then from the remaining $8$ toys, you select $4$ for Mike and one of the selections will be where Mike gets $T_1 - T_4$. Can you see how the groups are already ordered and why you should not multiply by $3!$?