Counting problem - I seem to be counting double If we have a group of $10$ men, and $4$ women, and we want to separate these $14$ people to $2$ groups of $7$ such that each group has at least $1$ women, in how many different ways can we achieve this?
My solution is: We can look at just one group, since that once we chose for the first group, we automatically have chosen for the second group, so we have $14$ people and we want to choose $7$ people such that there is at least $1$ women, so we either have $1$ women and $6$ men, or $2$ women and $5$ men, or $3$ women and $4$ men. That's:
$\binom{4}{1}\binom{10}{6}+\binom{4}{2}\binom{10}{5}+\binom{4}{3}\binom{10}{4}=3192$
So I'd say we have $3192$ ways to do this. However, the booklet says the correct answer is half of that,$1596$. I seem to be counting double. Where is my mistake?

Hint:
Suppose you have the people in a line, the women and the man. Then, you label each person with a number $1$ or $0$. $0$ for the first group and $1$ for the second one.
So $\color{pink}{1000}\color{blue}{1111110000}$ is one valid labeling (where pink ones are women and blue ones are men). But suppose you have this labeling too $\color{pink}{0111}\color{blue}{0000001111}$. What is the difference between those two forms of grouping them?
