Count the number of inversions in anticommutativity wedge product Let $f\in \Lambda^p(E^*),g\in \Lambda^q(E^*)$, then the wedge product is anticommutative, that is $g\wedge f=(-1)^{pq}f\wedge g$. The proof of this depends on some combinatorical element that shows that the number of inversions of the permutation
$$S_{p+q}\ni\tau:\left\{\begin{align} 1&\longmapsto q+1 \\ \vdots\\
p&\longmapsto p+q \\ p+1 &\longmapsto 1 \\
\vdots \\
p+q&\longmapsto q\end{align}\right.$$
is $pq$, so that $\varepsilon (\tau)=(-1)^{pq}$.
In all the courses I took this was left as an exercise, which I usually skipped since I am not fond of combinatorics. In the differential geometry course I am taking, this shows up for the for the third time, and I tried to prove it, in vain.
I tried to compose $\tau$ in the general case as a product of transpositions, by composing by a transposition $(1,q+1)$ such that $(1,q+1)\circ \tau:1\mapsto 1$ and then by a transposition $(2,q+2)$ such that $(2,q+2)\circ (1,q+1)\circ\tau:1\mapsto 1,2\mapsto 2$ but since $p$ and $q$ can be arbitrary, this soon comes out to be an extreme mess. Is there some easy way I am failing to see? All help will be helpful.
 A: An inversion is a pair $(i,j)$ where $i<j$ but $\tau_i>\tau_j$. There are three cases:


*

*If $i<j\le p$, then $\tau_i=q+i$ and $\tau_j=q+j$, so $\tau_i<\tau_j$. This is not an inversion.

*If $i\le p$ and $j> p$, then $\tau_i=q+i$ and $\tau_j=j-p$. You can show that $\tau_I>\tau_j$, so this is an inversion.

*If $p<i<j$, this is not an inversion (why?).
You will see that the only inversions come from case $2$. The number of pairs $(i,j)$ in case $2$ is $pq$; there are $p$ choices of $i$, and for each choice of $i\in \{1,2,\dots,p\}$, and there are $q$ choices of $j\in \{p+1,p+2,\dots,p+q\}$. Therefore, the number of inversions is $pq$.
A: I know this is an old post, but I thought I would be useful if an additional solution is added. Once said that, another way of proving that
$$
\varepsilon ( \tau)=(-1)^{pq}
$$
is to show that $\tau$ is the product of $pq$ transpositions. In which case, it is enough to observe that
$$
\tau=(1 \hskip.2cm p+q) \cdots (1 \hskip.2cm q+1)(2 \hskip.2cm p+q) \cdots (2 \hskip.2cm q+1) \cdots (q \hskip.2cm p+q) \cdots (q \hskip.2cm q+1)
$$
