Ways of distributing $10$ toys such that exactly $2$ children don't get any toy. $10$ different toys are to be distributed among $10$ children.  Total no. of ways of distributing these toys such that exactly $2$ children do not get any toy is equal to?
My working: $10 \choose 2$ ways of choosing $2$ children who don't get toys and $10^8$ ways of distributing toys to $8$ children.
So total no. of ways $= {10 \choose 2} \cdot 10^8$  
Answer given behind: $$(10!)^2 \cdot \left(\frac{1}{3!2!7!} + \frac{1}{2^46!}\right)$$
I have no idea whatsoever how to get here and whats wrong in my reasoning. I would really appreciate it if someone could point me in the right direction.
 A: 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.
A: Your approach counts those arrangements in which fewer than eight children receive a toy.  Let's modify your approach accordingly.
There are $\binom{10}{2}$ ways of choosing two children to not receive a toy.  There are $8^{10}$ ways of distributing the ten toys among the remaining eight children since there are eight possible recipients for each of the ten toys.  However, this counts those distributions in which fewer than eight children receive a toy.  We must exclude those cases.
There are $\binom{8}{k}$ of excluding $k$ of those eight children from receiving a toy and $(8 - k)^{10}$ ways of distributing the ten toys to the remaining $8 - k$ children.  By the Inclusion-Exclusion Principle, the number of ways of distributing the ten toys to exactly eight children is 
$$\sum_{k = 0}^{8} (-1)^k\binom{8}{k}(8 - k)^{10}$$
Hence, the number of ways of distributing the ten toys so that exactly two of the children do not receive a toy is 
$$\binom{10}{2}\sum_{k = 0}^{8} (-1)^k\binom{8}{k}(8 - k)^{10}$$
A: I think we choose first two children which will not receive toys, then distribute 8 toys among the 8 remaining kids (there are $10\cdot 9 \cdots \cdot 3=10!/2!$ ways to do it) and finally we decide if choose one or two children for the remaining toys. So, the answer should be $$\binom{10}{2}\cdot \dfrac{10!}{2!}\cdot \left(\binom{8}{1}+\binom{8}{2}\right)$$
