I am trying to find a formula for $a_{N, \ell}$ defined as follows:
Given $2N$ objects initially paired in some way, $a_{N,\,\ell}$ is the number of ways to select $\ell$ pairs (i.e., $2\ell$ objects), so that no object in these pairs is with its original partner.
By the definition of deranged matchings, we should find
$$a_{N, N} = (-1)^{N} \sum_{k=0}^N (-1)^k \binom{N}{k} \frac{(2k)!}{2^k k!}$$
(which is just $a_N$ defined below), and one could reason that $a_{N, \,1} = 2N(2N-2)/2$. Moreover, a related problem tells me that these coefficients should satisfy the identity
$$ \binom{2N}{2k} \frac{(2k)!}{2^k k!} = \sum_{\ell=0}^{k} \binom{N}{m} a_{N- m, \,k-m}.$$
Does anyone know a formula (and corresponding proof for said formula) for $a_{N, \,\ell}$?
Context
In "Avoiding Your Spouse at a Bridge Party", Margolius considers the "Bridge Couples Problem":
Suppose $N$ married couples ($2N$ people) are invited to a bridge party. Bridge partners are chosen at random without regard to gender. What is the probability that no one will be paired with his or her spouse?
She finds that the answer is $a_N/\Omega_N$ where $\Omega_N = (2N)!/2^N N!$ is the number of unique ways to place $2N$ people in pairs, and $a_N$ is the number of "deranged matchings" of $2N$ people with partners other than their spouse. That is, given $2N$ objects initially paired in some way, $a_N$ is the number of ways to repair the objects so that no object is with its original pair. By the principle of inclusion-exclusion, Margolius finds
$$a_N = (-1)^{N} \sum_{k=0}^N (-1)^k \binom{N}{k} \frac{(2k)!}{2^k k!} $$
(first several values are listed in https://oeis.org/A053871). These coefficients satisfy the identity
$$ \frac{(2N)!}{2^N N!} = \sum_{\ell=0}^N \binom{N}{\ell} a_{\ell}. $$
I am seeking a generalization of $a_N$ to the case where we choose $\ell< N$ pairs.
I was going to post this question without an answer, but after trying to explain the problem I found a path to the solution through the principle of inclusion-exclusion. The answer is
$$a_{N,\,\ell} = \sum_{j=0}^{\ell} (-1)^{j} \binom{N}{j} \binom{2N-2j}{2\ell-2j} (2\ell-2j-1)!!.$$
These coefficients can also be expressed as the integral
$$ a_{N,\, \ell} = \frac{1}{\Gamma(N- \ell + 1/2)}\binom{N}{\ell} \int^{\infty}_{0} dt\, e^{-t}\, t^{N-\ell -1/2}\,(2t - 1)^{\ell}.$$
BUT, it would be beneficial if someone could construct an alternative proof of this result.