I want to ask if I can find a $1$ from each row such that for every two of $1$ they are not in the same column.
Put some $1$ into an $n\times n$ grid (every check has at most one $1$) such that the number of $1$ in each row and each column is $k \ (
1
$\begingroup$
combinatorics
-
1Sure, put them all down the main diagonal. Or do you want the number of different ways to do it? – 2012-09-26
-
0http://en.wikipedia.org/wiki/Permutation_matrix – 2012-09-26