Rearranging pairs of $n$ twins such that $k\leq n$ are mismatched Say we have $n$ initial pairs of people where the people within a pair are identical (i.e., identical twins) but people in different pairs are distinguishable. What is the number of ways
to form $k\leq n$ pairs such that none of these new pairs consists of twin pairs?
Motivation/Context
Problem is related to the statistical physics of the dimerization of biomolecules, in particular a collection of homodimer proteins which are identical within the dimer but distinguishable from proteins in other dimers. 

Related Problems
I have been able to solve two related problem using the principle of inclusion-exclusion. Here are the statements and answers for reference: 
Problem I: Say we have $n$ initial man-woman couples where all the men and all the women are distinguishable.  What is the number of ways
to form $k\leq n$ man-woman couples such that none of these new couples coincides with any initial pairings?   
Answer: The number of ways to form $k\leq n$ man-woman couples such that none is from the initial set of couples is
\begin{equation}
b_{n, k} = \sum_{m=0}^{k} (-1)^{m} \binom{n}{m} \binom{n-m}{k-m}^2 (k-m)!.
\end{equation}
Problem II: Say we have $n$ initial couples where the all $2n$ people are distinguishable. Assuming each person can couple with any other person, what is the number of ways
to form $k$ couples such that none of these new couples coincides with any initial pairings? 
Answer: The number of ways to form $k\leq n$ the man-woman couples such that none is from the initial set of couples is
\begin{equation}
a_{n, k} = \sum_{j=0}^{k}(-1)^{j} \binom{n}{j} \binom{2n-2j}{2k-2j} (2k-2j-1)!!. 
\label{eq:anl_fin}
\end{equation}

This is completely changed, so some comments no longer apply.
Suppose there are $p$ pairs and $2q$ singletons among the $2k$ dimers, so $p+q=k$.  Let $f(p,q)$ be the number of ways to pair them up.
Let there be $p$ blue vertices snd $2q$ red vertices.  Two vertices are joined if those dimers are paired in the new arrangement.  It is possible for two blue vertices to be joined by two edges.  Each red vertex has degree 1, each blue vertex has degree 2. The edges form loops within the blue vertices, and paths whose ends are red vertices.
Take any red vertex, and follow the path via blue vertices until it reaches another red vertex.
$$f(p,q)=(2q-1)\left(f(p,q-1)+pf(p-1,q-1)+p(p-1)f(p-2,q-1)+...+p!f(0,q-1)\right)$$
Now, with only blue vertices left, take any blue vertex and follow the loop until it returns.
$$f(p,0)=(p-1)f(p-2,0)+\frac12[(p-1)(p-2)f(p-3,0)+...+(p-1)!f(0,0)]$$
The half is because a loop is the same as its reverse.
Lastly, multiply $f(p,q)$ by $n\choose{p,2q}$ and sum over $p+q=k$.  
EDIT
$f(p,0)$, that is without singles, is A002137.  Then 
$$f(p,q)=p!(2q-1)!!\sum_{r=0}^p\frac{f(r,0)}{r!}{p+q-1-r\choose q-1}$$
