estimating permanents with positive real entries taken from a set with two elements Suppose n is a positive integer > 1 and a, b are positive real numbers. Suppose A is an n by n matrix with n entries = to a and with n2 – n entries = to b. Is it true that permanent(A) achieves a minimum when all of the a’s occur in a single row of A? Is it true that permanent(A) achieves a maximum when all of the a’s occur along the diagonal of A?
Consider a similar min/max question: Suppose n is a positive integer > 1 and x1, …, xn are positive real numbers. Suppose B is an n by n matrix with n entries = to xk for k=1, … , n. Is it true that permanent(B) achieves a minimum when all of the xk’s occur in row k of B, for k=1, …, n? Is it true that permanent(B) achieves a maximum when B is a circulant matrix?
 A: The first statement (two values) is true, but the second is false.
Here's a proof of the first statement.
We need some basic facts about the permanent: it is multilinear (linear in each row and column), admits expansion by "minors," and is monotone in each entry.  Suppose the $a$'s in $A$ are not all in different rows and columns.  Then there must be a row or column containing no $a$'s (say a row), and a second row containing at least two $a$'s.  We will swap an $a$ from the second row with a $b$ in the first row, and show that the permanent increases (provided $a\ne b$).
Without loss, we're dealing with the first two rows, and the second row has an $a$ in the first column, so the picture is as follows:
$$A = \begin{pmatrix}
b &  {\bf v}\\
a &  {\bf w}\\
\vdots & B
\end{pmatrix}
$$
where $\bf v$ consists entirely of $b$'s, $\bf w$ contains at least one $a$, and $B$ is an $(n-2)\times(n-1)$ matrix.
Let $A'$ be the new matrix obtained by swapping the $a$ and $b$.  Expanding the permanents down the first column, and taking the difference, we get
$${\rm per} A' - {\rm per}A = (a-b)({\rm per}\binom{\bf w}B -{\rm per}\binom{\bf v}B).$$
If $a>b$, the first factor is positive, but so is the second by monotonicity of the permanent, as the components of $\bf w$ dominate those of $\bf v$.  Similarly, if $a<b$, both factors are negative, so in either case the product is positive, so we conclude
${\rm per} A' > {\rm per} A$.  Thus the permanent is maximized when the $a$ entries are in distinct rows and columns; this gives the same permanent as when the $a$ entries lie on the diagonal.  Walking the argument backwards, you decrease the permanent by introducing rows with all $b$'s, so the minimum occurs when all the $a$'s are in a single row. $\square$
As for the second statement, here's a counterexample.  Note that if a group $G$ has elements $x_1,\ldots,x_n$, the multiplication table for $G$ is a Latin square, which will be a circulant matrix iff $G$ is a cyclic group.  The set $\{0,1,\dots,7\}$ can be given a number of group structures; two natural ones are $(\Bbb Z/2\Bbb Z)^3$ (bitwise exclusive-or) and $\Bbb Z/8\Bbb Z$ (addition mod $8$).  Adding $1$ to the elements in both cases, to satisfy the technical condition that everything be positive, we get the two matrices
$$\left(
\begin{array}{cccccccc}
 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\
 2 & 1 & 4 & 3 & 6 & 5 & 8 & 7 \\
 3 & 4 & 1 & 2 & 7 & 8 & 5 & 6 \\
 4 & 3 & 2 & 1 & 8 & 7 & 6 & 5 \\
 5 & 6 & 7 & 8 & 1 & 2 & 3 & 4 \\
 6 & 5 & 8 & 7 & 2 & 1 & 4 & 3 \\
 7 & 8 & 5 & 6 & 3 & 4 & 1 & 2 \\
 8 & 7 & 6 & 5 & 4 & 3 & 2 & 1 \\
\end{array}
\right)
{\rm\ and\ }
\left(
\begin{array}{cccccccc}
 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\
 2 & 3 & 4 & 5 & 6 & 7 & 8 & 1 \\
 3 & 4 & 5 & 6 & 7 & 8 & 1 & 2 \\
 4 & 5 & 6 & 7 & 8 & 1 & 2 & 3 \\
 5 & 6 & 7 & 8 & 1 & 2 & 3 & 4 \\
 6 & 7 & 8 & 1 & 2 & 3 & 4 & 5 \\
 7 & 8 & 1 & 2 & 3 & 4 & 5 & 6 \\
 8 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\
\end{array}
\right),
$$
whose permanents are $7976955456$ and $7828309568$ respectively.
