$10$ 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.
Best Answer
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.