Q:

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

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

My try follows 
Choose  $3$ rows out of $4$ by $4C3= 4$
first ball has $4$ choices 
Second ball  has $3$ choices 
Third ball has $2$ choices 
So total number = $4×4×3×2=96$
Is my work right? 

A:

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.

