Consider the q-binomial coefficient $\binom{n}{k}_q$.

One combinatorial way to define it is as follows. Let $W_{n,k}$ be the set of binary words of length $n$ with $(n-k)$ 0's and $k$ 1's. An inversion in such a word $w=w_1\ldots w_n$ is a pair of indices $1 \leq i < j \leq n$ with $w_i > w_j$. Let $\mathrm{inv}(w)$ denote the number of inversions of $w$.

Then $\binom{n}{k}_q=\sum_{w\in W_{n,k}} q^{\mathrm{inv}(w)}$.

I am interested in plugging $q=-1$ into the binomial coefficient. For a word $w$, let $\mathrm{rev}(w)$ denote the reverse of $w$. So $w$ with $\mathrm{rev}(w) = w$ are the palindromes.

**Theorem**: $\binom{n}{k}_{q=-1} = \#\{w \in W_{n,k}\colon w = \mathrm{rev}(w)\}$.

It is possible to prove this theorem in an ugly way by evaluating $\binom{n}{k}_{q=-1}$ directly using the product formula $\binom{n}{k}_{q} = \frac{(1-q^n)(1-q^{n-1})\cdots(1-q^{n-k+1})}{(1-q^k)(1-q^{k-1})\cdots(1-q)}$. It is also possible to prove this in an algebraic way by realizing $\mathrm{rev}$ as a linear map on the exterior product $\wedge^k(\mathbb{C}^n)$.

As an intermediary between the ugly and the algebraic proofs, we can look for a nice combinatorial proof. In the case when **$n=2m$ is even**, I know a nice combinatorial proof via a sign-reversing involution, which I now briefly describe.

Take a word $w=w_1\ldots w_n \in W_{n,k}$. If $w$ is a palindrome, set $\tau(w) = w$. Otherwise, find the smallest $j$ such that $w_{m+1-j} \neq w_{m+j}$, and let $\tau(w)$ be the result of swapping these two letters in $w$. Then $\tau\colon W_{n,k}\to W_{n,k}$ is an involution which changes the parity of the number of inversions of the words it acts nontrivially on, and whose fixed points are precisely the palindromes (which all have an even number of inversions).

**Question**: Can this sign-reversing involution be extended to work in the case of $n$ odd?