How many ways can we place $3$ balls in $4×4$ grid such that no $2$ balls in the same row or column?

Yes, the answer is $$\binom{4}{3}4!=96$$ for precisely the reasons you describe (if the balls are indistinguishable).

You can choose a row and a column as undesirable in $4\cdot 4=16$ ways. Then you can select the three balls from the remaining array in as many ways as a $3\times3$ determinant has terms, namely $6$. Hence there are 96 admissible selections in total.