Distinguishable and indistinguishable objects in distinguishable containers My question is rather simple, but I can't seem to figure out how to provide an answer.
There are 5 distinguishable toys and 7 indistinguishable sweets that we try to give to 4 distinguishable children. How many ways are there to do that, so that every child gets at least one thing?
I could assign an object for every child first and then try to calculate for how many ways it's possible to assign the rest but that seems tricky as I would have to consider lots of cases. Actually all I can think of gets complicated quickly...
I did my research and I know there are similar questions, but I couldn't find any which would account for two type of objects (distinguishable and not) at once.
 A: For any event $E$ let $n\left(E\right)$ denote the number of distributions
such that $E$ occurs.
For $i=1,2,3,4$ let $E_{i}$ denote the event that child $i$ receives
no thing. 
Then to be found is $4^{5}\binom{7+3}{3}-n\left(E_{1}\cup E_{2}\cup E_{3}\cup E_{4}\right)$.
Here $4^{5}\binom{7+3}{3}$ equals the number of distributions
without regarding the condition that every child receives a thing.
Factor $4^{5}$ is linked to the distinghuishable toys and factor
$\binom{7+3}{3}$ to the indistinguishable sweets.
With inclusion/exclusion and symmetry we find: $$4^{5}\binom{7+3}{3}-n\left(E_{1}\cup E_{2}\cup E_{3}\cup E_{4}\right)=$$$$4^{5}\binom{7+3}{3}-4n\left(E_{1}\right)+6n\left(E_{1}\cap E_{2}\right)-4n(E_{1}\cap E_{2}\cap E_{3})=$$$$4^{5}\binom{7+3}{3}-4\times3^{5}\binom{7+2}{2}+6\times2^{5}\binom{7+1}{1}-4\times1^{5}\binom{7+0}{0}$$
If I made no mistakes then we end up with $89420$ distributions.
A: Choose $k$ of the children to give at least one toy to. By inclusion-exclusion, you can distribute the toys to them in
$$
\sum_{j=0}^k(-1)^j\binom kj(k-j)^5
$$
ways. Then give a sweet to each of the children who didn't get a toy, and distribute the remaining $7-(4-k)=k+3$ sweets to all $4$ children, in $\binom{k+6}3$ different ways. In total, this yields
$$
\sum_{k=1}^4\binom4k\binom{k+6}3\sum_{j=0}^k(-1)^j\binom kj(k-j)^5=89420\;.
$$
