Let $S\_n$ be the group of $n$-permutations. Denote the number of inversions of $\sigma\in S\_n$ by $\ell(\sigma)$.



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> **QUESTION.** Assume $n>2$. Does this cancellation property hold true?
>  $$\sum\_{\sigma\in S\_n}(-1)^{\ell(\sigma)}\sum\_{i=1}^ni(i-\sigma(i))=0.$$
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A cancellation property for permutations?
Let $n$ be some integer greater than 2. Since the number of even and odd permutations in $S\_n$ is the same we have $\sum\_{\sigma\in S\_{n}}(-1)^{\ell(\sigma)}=0$ therefore the contribution of $\sum\_{\sigma\in S\_{n}}(-1)^{\ell(\sigma)}\left(\sum\_{i=1}^n i^2\right)$ is zero. It remains to show that
$$\sum\_{\sigma\in S\_{n}}(-1)^{\ell(\sigma)}\sum\_{i=1}^n i\sigma(i)=0.$$
Notice that if we write $P(x)=\det\left(x^{ij}\right)\_{i,j=1}^n$ then this sum is simply $P'(1)$. However the order of vanishing of $P$ at $1$ is $\binom{n}{2}$ (notice that the matrix is pretty much a Vandermonde matrix) and this is greater than $1$ since $n>2$, therefore $P'(1)=0$.


