Distinguishable and indistinguishable objects in distinguishable containers

Question

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.

Answer 1

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.

Answer 2

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

$$ \sum_{k=1}^4\binom4k\binom{k+6}3\sum_{j=0}^k(-1)^j\binom kj(k-j)^5=89420\;. $$