An upper bound for the permanent of a $3$-dimensional $(0,1)$-matrix

Abstract:

Let $A = ({a_{ijk}})$ be a $3$-dimensional matrix of order $n$. The permanent of $A$ is defined by \[ \operatorname {per} A = \sum \limits _{\sigma ,\tau \in {S_n}} {\prod \limits _{i = 1}^n {{a_{i\sigma (i)\tau (i)}},} } \] where ${S_n}$ is the symmetric group on $\{ 1,2, \ldots ,n\}$. Suppose that $A$ is a (0,1)-matrix and that ${r_i} = \sum \nolimits _{j,k = 1}^n {{a_{ijk}}} {\text { for }}i = 1,2, \ldots ,n$. In this paper it is shown that per $A \leq \prod \nolimits _{i = 1}^n {{r_i}{!^{1/{r_i}}}.}$ A similar bound is then obtained for a second function, the $2$-permanent of a $3$-dimensional matrix, that is another analogue of the permanent of an ordinary ($2$-dimensional) matrix.