The problem:
< Find the base cases and a recursive step to the number of possibilities to place n different balls in r different cells when each cell must contain at least one ball.
I have two ideas that give different results and can't understand why one of them does not yield the correct answer.
The first solution is to the divide the problem to two different scenarios. In the first scenario, there is a cell which contains exactly one ball. In the other one, there is not.
So that gives: $T(n,r)=r(T(n-1,r-1)+T(n-1,r))$
In the first solution I again divide the problem to two different scenarios. In the first one, there is a single ball in cell number 1. In the other one, there are at least two.
That gives: $T(n,r)=n(T(n-1,r-1)+T(n-1,r))$
I can't understand where is my mistake. In either case there are two options - There is some cell that contains exactly one ball of there is not.
Obviously something is wrong, but what?
Your main problem is that you need make sure there is an exact match between some scenario and a 'recursive step'. I'll try to show where you made a mistake and how to argue about the correct solution.
The exact match between a scenario and a recursive step fails already for your first solution/first scenario: If you have a cell with exactly one ball in it, then you have $r-1$ remaining cells that contain $n-1$ balls, each cell also containg at least one ball. That's how you get the term $T(n-1,r-1)$. The factor $r$ comes from the fact that any of the $r$ cells could be the one with exactly one ball in it.
The problem is in the formulation "the one with ...", because it is easily possible that there are multiple such cells. In that case, you count the same solution multiple times!
Example: $n=4,r=3$. I put the elements of a cell in square brackets ([]), so one possible solution is
$$[1], [2], [3,4].$$
That means that ball $1$ is in the first cell, ball $2$ in the second cell and balls $3$ and $4$ in the third. In your first solution, you count this solution twice for the first scenario! It happens when you consider the first cell as the one with exactly one ball and take ($[2], [3,4]$) as one possible solution out of $T(3,2)$ and when you consider the second cell as the one with exactly one ball and take ($[1], [3,4]$) as one possible solution out of $T(3,2)$.
A similar problem exists with the second solution/second scenario: If you know that the first cell contains at least 2 balls, you somehow remove a ball from the first cell and arrive at the recursive step $T(n-1,r)$. But since you have multiple balls in that cell, you don't know which to remove and are again overcounting!
Example: $n=4,r=3$. All possible solutions for the second scenario of the second solutions are
$$([1,2], [3], [4]), ([1,2], [4], [3]), ([1,3], [2], [4]), ([1,3], [4], [2]), ([1,4], [2], [3]), ([1,4], [3], [2]), ([2,3], [1], [4]), ([2,3], [4], [1]), ([2,4], [1], [3]), ([2,4], [3], [1]), ([3,4], [1], [2]), ([3,4], [2], [1])$$
which makes $12$ alltogher. But since $T(3,3)=3!=6$, we get $12 \neq 4\times T(3,3)=24$. You overcount by a factor of $2$, since you get the solution $([1,2], [3], [4])$ (and every other one) twice: First by removing the $1$ from the first cell and counting $([2], [3], [4])$ from $T(3,3)$, then again by removing the $2$ from the first cell and counting $([1], [3], [4])$ from $T(3,3)$.
Since you don't know how many balls cell 1 contains, this approach doesn't work (at least not in one immediate step).
The correct solution is actually your first formula, just the reasoning needs to be different.
You have two scenarios: Is the ball number $n$ alone in a cell (scenario 1) or not (scenario 2)? We call the number of solutions $T_1(n,r)$ and $T_2(n,r)$, resp.
In the first scenario, there are $r$ cells that could (alone) contain ball $n$, and all the other $r-1$ cells contain the remaining $n-1$ balls, and each cell contains at least one ball, so we get
$$T_1(n,r)=rT(n-1,r-1)$$
Note that there is now no ambiguity which cell to select, ball $n$ is in exactly one cell.
In scenario 2, we can remove the ball $n$ and still get a valid distribution of $n-1$ balls into $r$ cells (valid meaning each cell contains at least one ball). Any such solution from $T(n-1,r)$ yields exactly $r$ solutions from $T_2(n,r)$: You can put the 'missing' ball $n$ in each of the $r$ cells. Again, there is no possibility of a solution from $T_2(n,r)$ being counted twice this way, because it is uniquely determined from which cell the ball $n$ has to be removed.
So we get
$$T_2(n,r)=rT(n-1,r)$$
and finally
$$T(n,r)=r(T(n-1,r-1) + T(n-1,r))$$