Creating two groups where two people can't be in the same group. If you have a group of $12$ men and $10$ women, and you need to make two groups - one with $6$ people and the other with $9$, the ways to form such groups would be
$${22 \choose 6}\cdot {16 \choose 9}$$
Since I guess the order doesn't matter.
But then, there is another constraint: Bob (male) and Hilda (female) cannot be in the same group.
For the time being, maybe I can remove them from the whole thing, and make one group with $5$ people and another with $8$ without them:
$${20 \choose 5}\cdot {15 \choose 8}$$
There are now $7$ people left, plus Bob and Hilda we got $9$. There is one spot free in the first group and another one in the second.
If Hilda is placed in the first group, then it is guaranteed Bob won't be in it (because it's full), so he would end up in the second group and viceversa.
So I guess the answer would be like this?
$${20 \choose 5}\cdot {15 \choose 8} \cdot 9 \cdot 8$$

In the initial case, you do not distinguish between men and women.  So effectively, you have $22$ distinct people to make up two groups one of which has $6$ members and the other has $9$ members.  Let's assume we have assembled all possible groups of $6$.  For each group of $6$, there would be $n$ sets of 9 elements from the remaining $22 - 6 = 16$.  This means $\binom{22 - 6}{9}$.  Which indeed gives your initial calculation: $\binom{22}{6}\cdot\binom{22 - 6}{9} = \binom{22}{6}\cdot\binom{16}{9}$.
Now, as you say, lets take these two individuals out.  They must always be on opposite teams.  The fact that one appears to be male and one appears to be female is irrelevant.  If we choose a set of, instead of $6$ and $9$ members, $5$ and $8$ members, from the total group of $22 - 2 = 20$ total members (the two excluded...).
This gives:
$$
\binom{20}{5}\cdot\binom{15}{8}
$$
However, for each of these groups there is $\binom{2}{1} = 2$ ways to put these two constrained members.  Since one group has $6$ members and the other $9$, each of these groups is unique!  Therefore switching which group Bob is in and which group Hilda is in does create two distinct groupings.
I do not believe it should be $9\cdot8$, rather:
$$
\binom{20}{5}\cdot\binom{15}{8} \cdot \binom{2}{1}
$$
In general for dividing into two regions of size $w$ and $h$ from a number of values numerating $N \geq w + h$.  For the number of unique set of two groups of size $w$ and $h$:
$$
n = \binom{N}{w}\cdot\binom{N - w}{h} = \binom{N}{h}\cdot\binom{N - h}{w}
$$
If we want to exclude two from being together we simply take them out--as you prescribed:
\begin{align}
n_{\text{divorce}} =& \binom{N - 2}{w - 1}\cdot\binom{N - 2 - (w - 1)}{h - 1}\cdot\binom{2}{1} \\
=& 2*\binom{N - 2}{w - 1}\cdot\binom{N - w  -1}{h - 1} \\
=& 2*\binom{N - 2}{h - 1}\cdot\binom{N - h - 1}{w - 1}
\end{align}
For your case this is:
\begin{align}
N =& 22\\
w =& 6 \\
h = & 9 \\
n =& 2*\binom{20}{5}\cdot\binom{22 - 6 - 1}{8} \\
=& 2*\binom{20}{5}\cdot\binom{16 - 1}{8} \\
=& 2*\binom{20}{5}\cdot\binom{15}{8} \\ \\
=& 2*\binom{20}{8}\cdot\binom{22 - 9 - 1}{5} \\
=& 2*\binom{20}{8}\cdot\binom{13 - 1}{5} \\
=& 2*\binom{20}{8}\cdot\binom{12}{5}
\end{align}
There is one more thing we should think about.  The above constraint is actually that both Bob and Hilda will always be chosen.  Assuming the only actual constraint is that they are not in the same group, then we also need to consider when exactly one of them is chosen and when neither of them is chosen.
This would mean you get:
$$
2*\binom{20}{8}\cdot\binom{12}{5} + 2*\binom{21}{9}\cdot\binom{12}{6} + \binom{20}{9}\cdot\binom{11}{6}
$$
