Distribute 5 distinguished balls to 3 undistinguished cell On a first glimpse it looks a very easy one, but I find that a bit odd.
Assuming we have 5 distinguished balls, and 3 undistinguished cells, in how many ways can we distribute the balls in the cells, when we have at least 1 ball in each? 
My first guess was to distribute the first 3 this way:
$$ \binom{5}{1,1,1} = \frac{5!}{1!} \cdot \frac{1}{3!} $$ 
I divided by $ 3! $ since the cells are undistinguished.
Then, For each of the remaining balls we have 3 options each, then it should be $ 3^2 $.
To sum up, it's 29.
But the answer is actually 25 :(
What am I doing wrong?
edit: I know that since this case is very simple I have 2 options to divide them: 3,1,1 and 2,2,1
But I'm actually interested in the general idea (that also works for big or abstract numbers).
Thanks

The numbers are small, so one can use cases: (i) $1,1,3$, and (ii) $1,2,2$.
Case (i) The "team of $3$" can be chosen in $\binom{5}{3}$ ways, and now we have no further choices.
Case (ii) The lonely one can be chosen in $\binom{5}{1}$ ways. For every such choice, the person from the remaining $4$ with the lowest student number can choose her partner in $\binom{3}{1}$ ways, for a total of $\binom{5}{1}\binom{3}{1}$.
Edit: The question was modified, and asks for information about larger numbers. The number of ways to partition a set of $n$ distinct objects into $k$ non-empty subsets is the Stirling Number of the Second Kind $S(n,k)$ There is no nice closed form for these, but there are useful recurrences. (The Stirling numbers of the second kind occur so often in combinatorics that some consider $S(n,k)$ to be a closed form.)
