Leibniz formula and determinants Let $A \in M_n(\mathbb C)$, then
$$\det(A) = \sum_{\sigma \in S_n} \mbox{sign}(\sigma) a_{1\sigma(1)} a_{2\sigma(2)} \cdots a_{n\sigma(n)} = \sum_{\sigma \in S_n}\prod_{i=1}^n a_{1\sigma(i)}$$
I looked at different resources to try and understand how the formula works, but it hasn't really clicked with me yet. I even read through Understanding the Leibniz formula for determinants and it helped a bit, but I still need some clarification. What throws me off the most are the subscripts for the elements in $A$, and I don't understand how you know which elements to multiply together.
If anyone can break it down into simpler terms, that would be greatly appreciated. Thanks!

It's not as bad as it seems! It means pick one element from each row of A such that no two elements are in the same column, then multiply them together. Add all such possible products together, with some plus and some minus. You put a plus if you chose the diagonal or something which can be placed on the diagonal by interchanging an even number of columns, and a minus if you have to interchange an odd number of columns to place your set on the diagonal.
