You are trying to count the number of committees with at least one man. What you are counting with the $\binom{9}{1}\binom{19}{6}$ is something else.
Imagine (if one can still imagine such a thing) that you are trying to count the number of ways of forming a committee of $7$ if the rules say that the Chair must be male, with the sexes of the rest of the people unspecified. Then $\binom{9}{1}\binom{19}{6}$ would be the right answer.
But with an undifferentiated committee, we are overcounting by quite a bit. What you are counting with your $\binom{9}{1}\binom{19}{6}$ is the set of all ordered pairs $(x,y)$, where $x$ is any man, and $y$ is any bag of $6$ people that does not contain $x$.
So for example, $x=$ Charlie, $y=\{$Dave, Bob, Mary, Mary-Ann, Jane, Janet$\}$ is one of the ordered pairs $(x,y)$ counted in your $\binom{9}{1}\binom{19}{6}$, but so is $x=$ Dave, $y=\{$Charlie, Bob, Mary, Mary-Ann, Jane, Janet$\}$. However, these result in the same committee of $7$.
And it is messy to compensate for the overcounting, because the amount of overcounting depends on the number of men chosen.