In how many ways can $5$ identical balls be placed in the cells of a $3 \times 3$ grid such that each row contains at least one ball? In how many ways can $5$ identical balls be placed in the cells of a $3 \times 3$ grid such that each row contains at least 1 ball?

I proceeded like this- In the first row choosing one cell out of $3$ is $3\choose1$, similarly $3\choose1$ for choosing one cell from 2nd row and third row. Now from remaining $6$ cells we can place remaining $2$ balls as $6\choose2$.
Thus the solution should be $\binom31\cdot\binom31\cdot\binom31\cdot\binom62$. 
But it isn't the right answer. Could anybody help me out? 
The right answer is 108. 

Using the fact that at most 1 row can be empty, subtract from the total ${9\choose 5}$ = 126 ways,
ways in which one row is full, one empty, and one with just 1 vacant slot.
There are 3! ways to permute the rows, and 3 ways to position the 1 vacant slot
Thus $126 - 3!\cdot 3 = 108$
