Extremly simple combinatorics - divide to groups We have a group of $10$ men and $4$ women, we want to divide this group into two groups of $7$ such that each of those groups has at least $1$ woman.
What I did:
I actually solved this in two directions, both of them are wrong, I'd like to know why.
First direction - First we choose a woman for group $A$, we have $4$ options for that. Then we choose a woman for group $B$, we have $3$ options for that. Now we have $12$ people that we need to divide into $2$ groups of $6$. We have $\begin{pmatrix} 12\\ 6\end{pmatrix}$ options for that. Overall thats $4 \times 3\times \begin{pmatrix} 12\\ 6\end{pmatrix} = 11088$
Second direction - Lets count all the number of groupings to groups of $7$ and subtract the ones where theres a group with no women.
Number of overall groupings - $\begin{pmatrix}14 \\ 7 \end{pmatrix}$. We have 14 people, we need to divide them into $2$ groups of $7$.
Number of groupings where theres a group without a woman - $\begin{pmatrix}10 \\ 7\end{pmatrix}$. Choose how to group the $10$ men only.
$\begin{pmatrix}14 \\ 7 \end{pmatrix} - \begin{pmatrix}10 \\ 7\end{pmatrix}=3432-120 = 3312$
Correct answer - $1596$.
What? How? Why? I'd like to know where my logic fails.
 A: They want unlabelled groups, and you have been thinking in terms of labelled groups. This is shown by your use of the terms Group A and Group B.
The number of ways of dividing the $14$ people into unlabelled groups is $\frac{1}{2}\cdot \binom{14}{7}$, half the number of ways of dividing into labelled groups.
If you subtract the $\binom{10}{7}$ ways of dividing into two groups, one all male, you will get the correct answer. Note that $\binom{10}{7}$ counts the number of ways to divide into unlabelled groups, one all male. For there are $2\cdot \binom{10}{7}$ ways to divide into two labelled groups, one all male.
Remark: Your first way involved multiple counting.  You put one woman, say Alicia, into Group A. Later, you may have put Beti and Cecille into Group A. That gives the same grouping as putting Beti into Group A, and later Alicia and Cecille.
A: You're on the right track, but you're overcounting. In your first method, suppose that you choose woman $W_1$ to be in group $A$, and then later add women $W_2$ and $W_3$, say, to that group. The problem is that you're counting this as distinct from first choosing $W_2$ and then adding $W_1$ and $W_3$, or first $W_3$ and then $W_1$ and $W_2$.
In your second method, note that $14 \choose 7$ chooses a group of $7$ from a total of $14$, but in doing so you have also determined the other group of $7$ (i.e., the ones who weren't chosen). Therefore you should divide by $2$. (It might be easier to look at smaller case, e.g., splitting $4$ people into $2$ groups of $2$. There are $\frac12{4 \choose 2} = 3$ ways to do so.)
Continuing from the second method, put all $4$ women into one group and choose $3$ men to join them: there are $10 \choose 3$ ways to do so. The answer is therefore
$$\frac12{14 \choose 7} - {10 \choose 3}.$$
A: Let us use a bit more notation.  Let us refer to the men and women by number, $M_1,M_2,\dots,W_1,W_2,W_3,W_4$.
The reason why your logic fails for the first attempt was that the sequence of steps: (Pick which woman goes to group A)(Pick which woman goes to group B)(Pick six more people to go to group A) winds up having the woman picked in step one as special compared to the others.
The following two sequences of choices give the "same" result:  


*

*$W_1, W_3, (W_2,M_1,M_2,M_3,M_4,M_5)$

*$W_2, W_3, (W_1,M_1,M_2,M_3,M_4,M_5)$


In both cases, our group $A$ looks like $(W_1,W_2,M_1,M_2,M_3,M_4,M_5)$.  Remember that order within the group doesn't matter.
Your second interpretation is almost correct given that the two groups are distinguishable.  That is, there is a "Group $A$" and a "Group $B$."  However, we aren't told that.  We can assume then that the two groups are not labeled.
I did say almost correct.  Your mistake was in that you only removed the cases where group $A$ was full of guys.  You need to also remove the cases where group $B$ was full of guys, for a total of $\binom{14}{7}-2\cdot \binom{10}{7} = 3192$
In order to account for the fact that the two groups are unlabeled and indistinguishable, we note that by counting group $A$ as different from group $B$, we double counted each scenario, if we divide by two we take care of the double-counting, for a final answer of $\frac{3192}{2}=1596$
