Understanding Determinants and Permutations I have a $4 \times 4$ matrix with all diagonal elements zero. How many of the $24$ products are sure to be zero?
I need some ideas how to solve this kind of question. Is there any systematic way, so I could solve the problem for any $n \times n$ matrix?
Further is it possible to generalize the problem: 
Lets say I have a $n \times n$ matrix and some zero elements. The zero elements can be in the same row, column, different column and row or a combination. Is it possible to solve this problem?
I guess its all about understand how Permutations work.
I appreciate your support.
thanks.
 A: For your original question, I believe that @user8734617's comment will give you all of the information you need. For the generalization of the problem, I would have you consider the following: Given a single zero in an $n\times n$ matrix, exactly $(n-1)!$ of the products will vanish. If there are $k$ zeroes in the same row (or column), and no zeroes anywhere else in the matrix, then $k\times (n-1)!$ products will vanish. Now if not all of the zeroes belong to the same row or column, we will need to use inclusion-exclusion reasoning to determine the number of products that will vanish. Fortunately, the counting isn't too hard for these different cases (they should all be of the form $(n-j)!$ when we have $j$ zeros having all different row indices and all different column indices), but it might be difficult to keep track of all the possible sets you need to consider.
Let's take a simple example to illustrate: Consider $$A=\begin{bmatrix}1&1&0&0\\1&0&1&1\\1&1&1&1\\1&1&1&1\end{bmatrix},$$ where I've made all the nonzero entries equal to $1$ for simplicity. To count the number of vanishing products, note that we have $3$ zeroes which each would remove $3!$ products if it were on its own; from this, we must subtract the $2$ sets of $2!$ products we are double-counting (one coming from the $(1,3)-(2,2)$ pair of zeroes, and one coming from the $(1,4)-(2,2)$ pair of zeroes, since no product has both of the $(1,3)-(1,4)$ entries). This gives $3\times 3!-2\times 2!=18-4=14$ vanishing products, which can be verified by counting directly.
For a second example, consider $$A=\begin{bmatrix}1&1&0&1\\1&0&1&1\\0&1&1&1\\1&1&1&1\end{bmatrix}.$$ Proceeding as before, we get $3\times 3!-3\times 2!+1\times 1!=18-6+1=13$ vanishing products. Note that while there are the same number of zero entries as in the first example, in this case no pair of them are in the same row or column, which is why this results in a different formula (and different final number).
