My first answer was: $$\binom{m+n-1}{m}$$ But then since things are non-numbered (identical classrooms) we have to get rid of overcounted combinations, which I found to be $n!$ ($n$ permutations of classes), so the answer is: $$\binom{m+n-1}{m}\Huge / n!$$
Is my answer correct?
Edit: I'm having doubts now, maybe the answer is just $\binom{m}{n}$?
The answer is not $\binom{m+n-1}m$, which would be true if the people were identical and the classrooms distinct, nor is it that value divided by $n!$.
If classrooms cannot be empty, the answer is by definition given by a Stirling number: $S(m,n)$. If classrooms can be empty, this changes to a sum of Stirling numbers $\sum_{k=1}^nS(m,k)$.