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}
 A: Let's call the answer to your original problem $c_{n,k}$.  We can compute $c_{n,k}$ by dividing $a_{n,k}$ out by the symmetries given by swapping within each initial pair.  More precisely, in the context of Problem II, let's say the initial pairs of people were $\{x_1,y_1\},\dots,\{x_n,y_n\}$.  Problem II then counts the set $S_{n,k}$ of all sets of $k$ disjoint two-element subsets of $\{x_1,y_1,\dots,x_n,y_n\}$, such that none of the subsets are among the original pairs.  Your original problem then counts the quotient of $S_{n,k}$ by the action of the group $G=(\mathbb{Z}/2\mathbb{Z})^n$ where the $j$th coordinate says whether to swap $x_j$ and $y_j$.
So, we just have to figure out how large the stabilizers of this action are.  If $P\in S_{n,k}$, let its width be the number of different $j$ such that either $x_j$ or $y_j$ is in some element of $P$.  Let $S_{n,k,w}$ be the set of elements of $S_{n,k}$ of width $w$ and let $a_{n,k,w}=|S_{n,k,w}|$.  Note that the stabilizer any $P\in S_{n,k,w}$ has at least $2^{n-w}$ elements: for each of the $n-w$ pairs $\{x_j,y_j\}$ are disjoint from all elements of $P$, swapping $x_j$ and $y_j$ will not change $P$.  However, the stabilizer could be larger: there are distinct $i,j$ such that both $\{x_i,x_j\}$ and $\{y_i,y_j\}$ (or both $\{x_i,y_j\}$ and $\{y_i,x_j\}$) are in $P$, then simultaneously swapping $x_i$ with $y_i$ and $x_j$ with $y_j$ will not change $P$, increasing the size of the stabilizer by a factor of $2$.
Let us call such a pair $\{i,j\}$ a bad pair, and let $S_{n,k,w,\ell}$ denote the set of those elements of $S_{n,k,w}$ that have exactly $\ell$ bad pairs.  Then the stabilizer of the action of $G$ for each $P\in S_{n,k,w,\ell}$ has $2^{n-w+\ell}$ elements, so each orbit of $G$ on $S_{n,k,w,\ell}$ has $2^{w-\ell}$ elements.  Thus we have $$c_{n,k}=\sum_{\ell=0}^{\lfloor k/2\rfloor}\sum_{w=k}^{\min(n,2k-2\ell)}\frac{a_{n,k,w,\ell}}{2^{w-\ell}}$$ where $a_{n,k,w,\ell}=|S_{n,k,w,\ell}|$.
Of course, for this formula to be of any use, we still need to compute $a_{n,k,w,\ell}$. First, note that the number of ways to choose and rearrange $\ell$ bad pairs is $$\frac{n!}{(n-2\ell)!\ell!}.$$ This is because there are $\frac{n!}{(n-2\ell)!}$ ways to choose $2\ell$ ordered elements of $\{1,\dots,n\}$ to form the pairs, then you divide by $2^\ell$ since the order within each pair does not matter, and divide by $\ell!$ since the order of the $\ell$ pairs does not matter.  However, each bad pair can be rearranged in two different configurations ($\{x_i,x_j\}$ and $\{y_i,y_j\}$ or $\{x_i,y_j\}$ and $\{y_i,x_j\}$), so we multiply by $2^\ell$.
Now, given a specific rearrangement of $\ell$ pairs to be bad as above, the number of elements of $S_{n,k,w}$ for which exactly those pairs are bad is $a_{n-2\ell,k-2\ell,w-2\ell,0}$.  We can then count $a_{n-2\ell,k-2\ell,w-2\ell,0}$ by starting from $a_{n-2\ell,k-2\ell,w-2\ell}$ and then using inclusion-exclusion to exclude all cases with any bad pairs, similar to how you solved your Problems I and II.  This gives 
$$\begin{align*}
a_{n,k,\ell,w}&=\frac{n!}{(n-2\ell)!\ell!}\sum_{m=0}^{\lfloor k/2\rfloor-\ell}(-1)^m\frac{(n-2\ell)!}{(n-2\ell-2m)!m!}a_{n-2\ell-2m,k-2\ell-2m,w-2\ell-2m} \\
&=\frac{n!}{\ell!}\sum_{m=0}^{\lfloor k/2\rfloor-\ell}(-1)^m\frac{a_{n-2\ell-2m,k-2\ell-2m,w-2\ell-2m}}{(n-2\ell-2m)!m!}
\end{align*}$$
Now we just have to compute $a_{n,k,w}$.  We can do this by an inclusion-exclusion process similar to how you computed $a_{n,k}$.  First, note that the number of ways to pick $2k$ elements of $\{x_1,y_1,\dots,x_n,y_n\}$ which intersect $w$ of the original pairs is $$2^{2w-2k}\binom{n}{2w-2k}\binom{n-2w+2k}{2k-w}.$$  Here we are first picking $2w-2k$ elements of distinct original pairs, and then $2k-w$ more full pairs, to give a total of $(2w-2k)+2(2k-w)=2k$ elements spanning $(2w-2k)+(2k-w)=w$ different pairs.  Having picked these elements, there are then $(2k-1)!!$ ways to arrange them into new pairs, without regard for whether all the new pairs are mismatched.  Using inclusion-exclusion to remove the cases where some new pairs are not mismatched, we get $$a_{n,k,w}=\sum_{j=0}^{2k-w}(-1)^j\binom{n}{j}2^{2w-2k}\binom{n-j}{2w-2k}\binom{n-2w+2k-j}{2k-w-j}(2k-2j-1)!!.$$
Combining all these formulas gives the following rather horrendous formula for $c_{n,k}$ that does not fit on one line:
$$c_{n,k}=\sum_{\ell=0}^{\lfloor k/2\rfloor}\sum_{w=k}^{\min(n,2k-2\ell)}\sum_{m=0}^{\lfloor k/2\rfloor-\ell}\sum_{j=0}^{2k-w-2\ell-2m} \\ (-1)^{m+j} 2^{w-2k+\ell} \frac{n!(2k-4\ell-4m-2j-1)!!}{\ell!m!j!(2w-2k)!(2k-w-2\ell-2m-j)!(n-w)!}$$
A: 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 
$$f(p,q)=p!(2q-1)!!\sum_{r=0}^p\frac{f(r,0)}{r!}{p+q-1-r\choose q-1}$$
