TITLE: Can we count odd and even derangements nicely without taking a determinant?
QUESTION [11 upvotes]: It's not hard to see that 
$$\det \begin{pmatrix}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0 \end{pmatrix}$$
is equal to
#(even derangements on 3 elements) - #(odd derangements on 3 elements)
and similarly for larger n.  It's not hard to calculate this determinant by various methods, and together with the known expression for the total number of derangements on n elements this results in explicit expressions for the number of odd and even derangements on n elements.
Question: Is there any nice and fundamentally different way of getting at the numbers of odd an even derangements?
My motivation is that this would then provide an alternate method of calculating the determinant.  See:
Matrix with zeros on diagonal and ones in other places is invertible
which is the original motivation, goes over a few simple ways to calculate the determinant, and includes a full explanation of the identity I claim above.

REPLY [2 votes]: Perhaps this is not what you are looking for, but we can simplify the computation of the determinant to a $1\times1$ determinant, which is as close to not-a-determinant as a determinant gets.
In this answer, it is shown that
$$
\det(\lambda I_n-AB)=\lambda^{n-m}\det(\lambda I_m-BA)
$$
Using $\lambda=1$, $m=1$, $A=\left.\begin{bmatrix}1\\1\\1\\\vdots\\1\end{bmatrix}\right\}$${\scriptsize n\text{ tall}}$, and $B=\underbrace{\begin{bmatrix}1&1&1&\cdots&1\end{bmatrix}}_{\text{$n$ wide}}$,
we get
$$
(-1)^n\det\underbrace{\begin{bmatrix}
0&1&1&\cdots&1\\
1&0&1&\cdots&1\\
1&1&0&\cdots&1\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
1&1&1&\cdots&0
\end{bmatrix}}_{n\times n}
=\det(I_n-AB)=\det\underbrace{(1-BA)}_{1\times1}=1-n
$$
Therefore,
$$
\det\underbrace{\begin{bmatrix}
0&1&1&\cdots&1\\
1&0&1&\cdots&1\\
1&1&0&\cdots&1\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
1&1&1&\cdots&0
\end{bmatrix}}_{n\times n}
=(-1)^{n-1}(n-1)
$$