The parity of permutation $(b_1, ..., b_{n-k},a_1, ..., a_k)$ Let $k$ be an integer between $1$ and $n$.
Let $\sigma_1$ be the parity of  $(a_1, ... ,a_k, b_1, ... ,b_{n-k})$ as a permutation of ${\{1, ... n}\}$ . So $a_1, ... ,a_k, b_1, ... ,b_{n-k}$ are all distinct numbers between $1$ and $n$.
Let $\sigma_2$ be the parity of  $(b_1, ..., b_{n-k},a_1, ..., a_k)$ .
($\sigma_i = 1$ or $0$ depending on whether the permutation is odd or even.)
I just know the result that $ (-1)^{\sigma_1+\sigma_2} = (-1) ^{k(n-k)}$, but cannot think up a reason for it. The problem seems combinatoric, I think. Please share your ideas, Thanks!
 A: As nasekatnasushi mentioned, the parity of a permutation $p$ $\sigma(p)$ can be defined as the parity of $N(p)$, the number of inversions in $p$. That is, $\sigma(p)=1$ if $N(p)$ is odd while $\sigma(p)=0$ if $N(p)$ is even. We have $(-1)^{\sigma(p)}=(-1)^{N(p)}$.

There are three groups of inversions in $p_1=(a_1, \cdots ,a_k, b_1, \cdots ,b_{n-k})$.

*

*inversions among element $a_i$'s

*inversions among element $b_j$'s

*inversions between an element $a_i$ and an element $b_j$.

$$\sum_{1\le i_1<i_2\le k}[a_{i_1}{>}a_{i_2}] + \sum_{1\le j_1<j_2\le n-k}[b_{j_1}{>}b_{j_2}] +\sum_{1\le i\le k,\,1\le j\le n-k}[a_i{>}b_j].  $$
Similarly, there are three groups of inversions in $p_2=(b_1,\cdots ,b_{n-k}, a_1, \cdots, a_k)$.

*

*inversions among element $a_i$'s

*inversions among element $b_j$'s

*inversions between an element $a_i$ and an element $b_j$.

$$\sum_{1\le i_1<i_2\le k}[a_{i_1}{>}a_{i_2}] + \sum_{1\le j_1<j_2\le n-k}[b_{j_1}{>}b_{j_2}] +\sum_{1\le i\le k,\,1\le j\le n-k}[a_i{<}b_j].  $$
Since element $a_i$'s are ordered in $p_1$ the same as in $p_2$, the inversions among $a_i$'s in $p_1$ are the same as those in $p_2$. Hence all of them together do not affect the parity of the sum of the number of inversions in $p_1$ and that in $p_2$. Ditto for the inversions among $b_j$'s in $p_1$ and that in $p_2$.
Every pair of elements $(a_i, b_j)$ is an inversion in exactly one of $p_1$ and $p_2$.
Hence the parity of the sum of the number of inversions in $p_1$ and that in $p_2$ is the same as the parity of the number of pairs $(a_i,b_j)$, which is $k(n-k)$.
$$(-1)^{\sigma_1+\sigma_2}=(-1)^{N(p_1)+N(p_2)}=(-1)^{k(n-k)}.$$
