Evaluating the sum over all strings made of two anticommuting terms Given two anticommuting elements, $A$ and $B$, I aim at evaluating the sum over all strings of length $n$ multiplying exactly $k$ elements $A$ and $n-k$ elements $B$ (as we know, there are $\binom{n}{k}$ strings in total). In maths terms, I want to evaluate the constant $c$ in the expression
\begin{align*}
c\cdot A^k \, B^{n-k} &= \text{sum over all strings containing $k$ As and $n-k$ Bs}\\
&=\sum_{\sigma\in S_n} \frac{1}{k!(n-k)!}\sigma(A^k \, B^{n-k}),
\end{align*}
where $\sigma$ is a permutation in $S_n$.
The intermediate step (but maybe not necessary?) would be to calculate $p(\sigma)$ for
\begin{equation*}
p(\sigma) \cdot A^k \, B^{n-k} = \sigma(A^k \, B^{n-k})
\end{equation*}
which should be either $1$ or $-1$ depending on whether the number of adjacent transposition needed to turn the ordered string $A^k \, B^{n-k}$ into its permuted version under $\sigma$ is even or odd.
Is there a formula for $c$ (and $p$) in terms of $n$ and $k$?
 A: Let $C(n,k)$ denote the summation of all strings of $k$ $A$'s and $n-k$ $B$'s, so that $C(n,k)=c(n,k)A^kB^{n-k}$. Since
$$
\begin{align}
c(n,k)A^kB^{n-k}=C(n,k)
&=A\cdot C(n-1,k-1)+B\cdot C(n-1,k)\\
&= A\cdot c(n-1,k-1) A^{k-1}B^{n-k} +B\cdot c(n-1,k)\cdot A^{k}B^{n-k-1}\\
&=c(n-1,k-1) A^{k}B^{n-k} + c(n-1,k)\color{red}{(-1)^k}\cdot A^{k}B^{n-k}\\
&=[c(n-1,k-1)+(-1)^k c(n-1,k)] A^{k}B^{n-k}
\end{align}
$$
We conclude that
$$
c(n,k)=c(n-1,k-1)+(-1)^k c(n-1,k),
$$
so $c(n,k)$ satisfies a twisted version of Pascal's rule. Using this, together with the base cases $c(0,k)=1[k=0]$, you can prove by induction that
$$
c(n,k)=\begin{cases}
\binom{\lfloor n/2\rfloor }{\lfloor k/2\rfloor } & \text{if $n$ is odd or $k$ is even}
\\
0 & \text{if $n$ is even and $k$ is odd}
\end{cases}
$$
Putting this triangle of integers into OEIS, we find the entry https://oeis.org/A051159, which also gives this description for $c(n,k)$:

Coefficients in expansion of $(x + y)^n$ where $x$ and $y$ anticommute ($y x = -x y$), that is, $q$-binomial coefficients when $q = -1$.

That is, $c(n,k)=\binom{n}{k}_{-1}$.
