Let us first consider an $n\times n$ matrix with zero diagonal and $n(n-1)$ indeterminates. Every term (monomial) of the determinant corresponds to a permutation matrix $P$ with zero diagonal (i.e., an integer nonnegative matrix with trace 0 and row and column sums 1).

Let us now consider a symmetric $n\times n$ matrix with zero diagonal and $n(n-1)/2$ indeterminates $a_{ij}=a_{ji}$. Then every term corresponds to a permutation matrix $P$ as before, but the correspondence is not one-to-one, since symmetric elements are equal. We can get a one-to-one correspondence by associating the term to $P+P^T$ (the "symmetrized version" of $P$): an integer nonnegative matrix with trace 0 and row sums 2. The coefficient of the term is the $\det(P+P^T)$.

Now, in the Cayley-Menger determinant, the entries of the first row and column are not indeterminates but ones. We have to argue that this does not cause a "loss of information". In fact, the matrix $P+P^T$ is the adjacency matrix of a 2-regular undirected graph $G$ on $n$ vertices, without loops but potentially with multiple edges. This graph is a straightforward representation of every monomial in the determinant. Setting all variables $a_{12},a_{13},\ldots,a_{1n}$ to $1$ corresponds to eliminating the edges incident to vertex 1, resulting in a graph $G'$ on $n-1$ vertices. However, since $G$ was 2-regular without loops, we can uniquely identify the missing edges: they are incident to the vertices of degree less than 2 in $G'$.

The coefficient of the monomial is still the determinant of $P+P^T$, which equals $\pm2^k$, where $k$ is the number of cycles of length at least 3 in the graph $G$. The sign is the sign of the (more precisely, of any) permutation $P$.

Let us first consider an $n\times n$ matrix with zero diagonal and $n(n-1)$ indeterminates. Every term (monomial) of the determinant corresponds to a permutation matrix $P$ with zero diagonal (i.e., an integer nonnegative matrix with trace 0 and row and column sums 1).

Let us now consider a symmetric $n\times n$ matrix with zero diagonal and $n(n-1)/2$ indeterminates $a_{ij}=a_{ji}$. Then every term corresponds to a permutation matrix $P$ as before, but the correspondence is not one-to-one, since symmetric elements are equal. We can get a one-to-one correspondence by associating the term to $P+P^T$ (the "symmetrized version" of $P$): an integer nonnegative symmetric matrix with trace 0 and row sums 2. The coefficient of the term is $\det(P+P^T)$.

Now, in the Cayley-Menger determinant, the entries of the first row and column are not indeterminates but ones. We have to argue that this does not cause a "loss of information". In fact, the matrix $P+P^T$ is the adjacency matrix of a 2-regular undirected graph $G$ on $n$ vertices, without loops but potentially with multiple edges. This graph is a straightforward representation of every monomial in the determinant. Setting all variables $a_{12},a_{13},\ldots,a_{1n}$ to $1$ corresponds to eliminating the edges incident to vertex 1, resulting in a graph $G'$ on $n-1$ vertices. However, since $G$ was 2-regular without loops, we can uniquely identify the missing edges: they are incident to the vertices of degree less than 2 in $G'$.

The coefficient of the monomial is still the determinant of $P+P^T$, which equals $\pm2^k$, where $k$ is the number of cycles of length at least 3 in the graph $G$. The sign is the sign of the (more precisely, of any) permutation $P$.

Let us first consider an $n\times n$ matrix with zero diagonal and $n(n-1)$ indeterminates. Every term (monomial) of the determinant corresponds to a permutation matrix $P$ with zero diagonal (i.e., an integer nonnegative matrix with trace 0 and row and column sums 1).

Let us now consider a symmetric $n\times n$ matrix with zero diagonal and $n(n-1)/2$ indeterminates $a_{ij}=a_{ji}$. Then every term corresponds to a permutation matrix $P$ as before, but the correspondence is not one-to-one, since symmetric elements are equal. We can get a one-to-one correspondence by associating the term to $P+P^T$ (the "symmetrized version" of $P$): an integer nonnegative matrix with trace 0 and row sums 2. The coefficient of the term is the $\det(P+P^T)$.

Now, in the Cayley-Menger determinant, the entries of the first row and column are not indeterminates but ones. We have to argue that this does not cause a "loss of information". In fact, the matrix $P+P^T$ is the adjacency matrix of a 2-regular undirected graph $G$ on $n$ vertices, without loops but potentially with multiple edges. This graph is a straightforward representation of every monomial in the determinant. Setting all variables $a_{12},a_{13},\ldots,a_{1n}$ to 1 corresponds to eliminating the edges incident to vertex 1, resulting in a graph $G'$ on $n-1$ vertices. However, since $G$ was 2-regular, we can uniquely identify the missing edges: they are incident to the vertices of degree less than 2.

The coefficient of the monomial is still the determinant of $P+P^T$, which equals $2^k$, where $k$ is the number of cycles of length at least 3 in the graph $G$.

Let us first consider an $n\times n$ matrix with zero diagonal and $n(n-1)$ indeterminates. Every term (monomial) of the determinant corresponds to a permutation matrix $P$ with zero diagonal (i.e., an integer nonnegative matrix with trace 0 and row and column sums 1).

Let us now consider a symmetric $n\times n$ matrix with zero diagonal and $n(n-1)/2$ indeterminates $a_{ij}=a_{ji}$. Then every term corresponds to a permutation matrix $P$ as before, but the correspondence is not one-to-one, since symmetric elements are equal. We can get a one-to-one correspondence by associating the term to $P+P^T$ (the "symmetrized version" of $P$): an integer nonnegative matrix with trace 0 and row sums 2. The coefficient of the term is the $\det(P+P^T)$.

Now, in the Cayley-Menger determinant, the entries of the first row and column are not indeterminates but ones. We have to argue that this does not cause a "loss of information". In fact, the matrix $P+P^T$ is the adjacency matrix of a 2-regular undirected graph $G$ on $n$ vertices, without loops but potentially with multiple edges. This graph is a straightforward representation of every monomial in the determinant. Setting all variables $a_{12},a_{13},\ldots,a_{1n}$ to $1$ corresponds to eliminating the edges incident to vertex 1, resulting in a graph $G'$ on $n-1$ vertices. However, since $G$ was 2-regular without loops, we can uniquely identify the missing edges: they are incident to the vertices of degree less than 2 in $G'$.

The coefficient of the monomial is still the determinant of $P+P^T$, which equals $\pm2^k$, where $k$ is the number of cycles of length at least 3 in the graph $G$. The sign is the sign of the (more precisely, of any) permutation $P$.