$(a)$ If $a_{11}=a_{22}=a_{33}=0$, how many of the $6$ terms in $det A$ will be zero?
$(b)$ If $a_{11}=a_{22}=a_{33}=a_{44}=0$, how many of the $24$ products $a_{1j}a_{2k}a_{3l}a_{4m}$ are sure to be zero?
My question resolve around question $(b)$. I do not understand where $24$ products arise from. I am guessing that it is a $4 \times 4$ matrix (all I can mathematically related is $(16-4) \cdot 2 = 24$)?
In general, the number of products in the determinant of an $n \times n$ matrix is $n!$ (so in this case, we have $4! = 24$ terms). This can be seen by applying an induction argument to a cofactor expansion. Indeed, the number of terms when cofactor expanding along a row is $n$ times the number of terms in the determinant of a matrix of size $(n - 1) \times (n - 1)$.