The usual formula for determinant of an arbitrary matrix is given by: $$\mbox{det}A = \sum_{\sigma \in S_{n}}\epsilon_{\sigma}\prod_{k=1}^{n}a_{k,\sigma(k)}$$ where $\sum_{\sigma \in S_{n}}$ denotes the sum of every permutation $\sigma$ of the set $S_{n}:=\{1,...,n\}$ and $\epsilon_{\sigma}$ is the sign of the associate permutation $\sigma$.
Now, if $A = (a_{ij})$ be an $n\times n$ symmetric matrix, does $$\mbox{det}A = \frac{1}{n!}\sum_{\sigma \in S_{n}}\epsilon_{\sigma}\prod_{k=1}^{n-1}a_{\sigma(k)\sigma(k+1)}$$ hold?