For $k\in \mathbb{N}$, $A$, an $n$ x $n$ matrix has the following properties:
- Every entry belongs to the set $\{0,1,2,...,k \}$, and,
- Every individual row and column adds up to $k$.
Question: Can we always permute just the rows (or just the columns) of $A$ to obtain a matrix which has all non-zero diagonal entries?
Further, I encountered this question while solving the following recreational question on cards.
Say I have an usual pack of $52$ cards ($13$ cards of each type numbered from $1$ to $13$). In any arbitrary distribution of these cards into $13$ groups, can I always find an ordering of the groups such that group $i$ contains a card numbered $i$?
I observed that if I form a matrix with the mentioned properties, where $n=13$, and $k=4$, and the $A_{ij}$ entry is the number of $j$ numbered cards in $i^{th}$ group, then the question on the matrix property is an equivalent of the question on cards.
Finally, I have proved it is possible for any $k$, and $n=3$ and $4$. I would really appreciate further help. Thank you.