I am working on the following problem:
$ \bullet $ How many possibilities exist to assign $n$ people to $k$ rooms, when the rooms are different and there might be empty rooms.
My attempt (version $1$). The problem becomes quite easy when we look at it in the following way: we want to assign a room to each one of the $n$ people and each room might be assigned to $0,\dots,n$ people. With this in mind, denoting the set of people by $\{ P_1, \dots, P_n \}$, we have $k$ possibilites for each $P_i$ and then the answer comes given by $k^n.$
My attempt (verson $2$ - using Stirling numbers). Since the rooms are different, we know that the number $$ j! \, {n \brace j }$$ gives us the number of ways to assign $n$ people into $j$ different rooms. Furthermore, if some rooms might be empty, the number $$ \binom{k}{j}j! \, {n \brace j}$$ gives us the number of ways to assign $n$ people into $j$ different rooms, which are picked from $k$ total rooms (this means that $k-j$ rooms are empty, here we just picked the rooms that must be non empty). Then, for example, the number of ways to assing $n$ people into $1$ room (of the total $k$ rooms) is given by $$ \binom{k}{1} 1! \, {n \brace 1} $$ and so on. Then, this leads me to the conclusion that the answer we are looking for is $$ \sum_{j=1}^k \binom{k}{j}j! \, {n \brace j}. $$
Concerns. With further search, I found out the identity $$ k^n = \sum_{j=0}^\color{red}{n} \binom{k}{j}j! \, {n \brace j} = \sum_{j=1}^\color{red}{n} \binom{k}{j}j! \, {n \brace j}$$ holds for every $k \in \mathbb R.$ BUT my answer isn't exactly this. With the two resolutions above, I have shown that $$ k^n = \sum_{j=1}^\color{red}{k} \binom{k}{j}j! \, {n \brace j} $$ And I really can't see where I am messing this up $(?)$. Thanks for any help in advance.
This is all very nicely thought through – you were only missing a tiny detail in the end. Both the binomial coefficient $\binom kj$ and the Stirling number $\left\{n\atop k\right\}$ are zero if the lower index is greater than the upper index. So the sum really only runs up to $\min(k,n)$ in either case.