Both the toys and the children are distinguishable, so this is not a stars and bars problem: it would be a stars and bars problem if the toys were indistinguishable.
HINT: There are $n$ ways to choose the child who is to receive no toy. There are $\binom{n}2$ ways to form a pair of toys to be given to the lucky child, and there are $n-1$ ways to choose who is to get that pair. Now there are $n-2$ single toys to be distributed to the remaining $n-2$ children so that each gets one toy. Can you put the pieces together now to come up with the final result?
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
The solution is not correct. As an illustration, let us distribute $5$ toys instead. Let us solve this simpler problem using the argument of the post. That procedure would suggest a count of $\binom{5}{4}(4!)(4)=480$.
Now let us count the actual number of ways to distribute the $5$ toys. Some kid will get $2$ toys, the rest will get $1$ each. The lucky kid can be chosen in $4$ ways. For each of these ways, the toys she gets can be chosen in $\binom{5}{2}$ ways. And for each of these ways, the rest of the distribution can be done in $3!$ ways, for a total of $240$.
Remark: For $6$ toys, the answer turns out to be $1560$. Note that there is no simple "overcount ratio."
In general, the number of ways to distribute $n$ toys among $k$ children so that each gets at least one toy is $k!S(n,k)$, where $S(n,k)$ is the Stirling number of the second kind.
There are nice recurrences that make computing the Stirling numbers of the second kind not too hard, at least for small inputs. Also, a number of programs will do the job. Wolfram Alpha does it for free. Just type in $S2(6,4)$.
Added: There is a simple way to see that $5760$ cannot be correct. Let us count the number of ways to split the $6$ toys between the $4$ kids, with no restrictions. So one or more kids may get nothing. Line up the toys in ascending order of desirability. The worst toy can be assigned in $4$ ways. For each of these, the second worst can be assigned in $4$ ways, and so on, for a total of $4^6=4096$. This is less than $5760$!
This beginning, together with the Principle of Inclusion/Exclusion, can be used to count the number of ways to distribute the toys so that each kid gets at least one toy.
For a quick and dirty count for $6$ toys and $4$ kids, we use the method that was illustrated earlier for $5$ toys and $4$ kids.
Maybe (i) a kid gets $3$ toys, and the rest get $1$ each or (ii) two kids get $2$ toys each, and the rest get $1$ each.
Counting the patterns in (i) is easy. The lucky kid can be chosen in $4$ ways. For each way, the toys she gets can be chosen in $\binom{6}{3}$ ways. And now the remaining toys can be distributed in $(3)(2)(1)$ ways, for a total of $480$ ways of type (i).
Counting the patterns in (ii) is tricky, just like counting the number of "two pair" poker hands is tricky. The two lucky kids can be chosen in $\binom{4}{2}$ ways. For each of these ways, the toys for the younger of the two lucky kids can be chosen in $\binom{6}{2}$ ways. And now the toys for the other lucky kid can be chosen in $\binom{4}{2}$ ways. Finally, the rest of the toys can be assigned in $(2)(1)$ ways. The total number of ways of type (ii) is $1080$.
This gives $1560$ as the number of ways to do the distribution.