Let's call the women A, B, C, D, E, F, G, H, I and the men S, T, U, V, W, X, Y, Z. If you are going to choose $5$ of the women and then $7$ of the rest, here is one way to do it:
(1) choose the women A, B, C, D, E and then choose Z, Y, X, W, V, U, I.
Here is another way:
(2) choose the women A, B, C, D, I and then choose Z, Y, X, W, V, U, E.
But actually the results of these two choices are the same. So you have counted this possibility twice - in fact, if you think about it, more than twice. If you do all the calculations you should find that your second answer is greater than the correct answer.
If Order is Unimportant
The number of ways to choose $5$ letters (if their order is unimportant) is the coefficient of $x^5$ in
$$
\begin{align}
&\small\overbrace{(1+x)\vphantom{x^2}}^{1\text{ E}}
\overbrace{\left(1+x+x^2\right)}^{2\text{ D's}}
\overbrace{\left(1+x+x^2+x^3\right)}^{3\text{ C's}}
\overbrace{\left(1+x+x^2+x^3+x^4\right)}^{4\text{ B's}}
\overbrace{\left(1+x+x^2+x^3+x^4+x^5\right)}^{5\text{ A's}}\\
&=\frac{1-x^2}{1-x}\frac{1-x^3}{1-x}\frac{1-x^4}{1-x}\frac{1-x^5}{1-x}\frac{1-x^6}{1-x}\\
&=\frac{1-x^2-x^3-x^4+O\left(x^7\right)}{(1-x)^5}\\[3pt]
&=\small\left[1-x^2-x^3-x^4+O\!\left(x^7\right)\right]\!\left[1+5x+15x^2+35x^3+70x^4+126x^5+210x^6+O\!\left(x^7\right)\right]\\[9pt]
&=1+5x+14x^2+29x^3+49x^4+71x^5+90x^6+O\!\left(x^7\right)\tag{1}
\end{align}
$$
where we used the Binomial Theorem for $(1-x)^{-5}$ above.
The coefficient of $x^5$ in $(1)$ is $71$.
If Order is Important
If the order of the letters is important, we can compute the exponential generating function with
$$
\begin{align}
&\small\overbrace{(1+x)\vphantom{\frac{x^2}{2!}}}^{1\text{ E}}
\overbrace{\left(1{+}x{+}\frac{x^2}{2!}\right)}^{2\text{ D's}}
\overbrace{\!\left(1{+}x{+}\frac{x^2}{2!}{+}\frac{x^3}{3!}\right)}^{3\text{ C's}}
\overbrace{\!\left(1{+}x{+}\frac{x^2}{2!}{+}\frac{x^3}{3}{+}\frac{x^4}{4!}\right)}^{4\text{ B's}}
\overbrace{\!\left(1{+}x{+}\frac{x^2}{2!}{+}\frac{x^3}{3!}{+}\frac{x^4}{4!}{+}\frac{x^5}{5!}\right)}^{5\text{ A's}}\\[3pt]
&=\small1+5x+24\frac{x^2}{2!}+111\frac{x^3}{3!}+494\frac{x^4}{4!}+2111\frac{x^5}{5!}+8634\frac{x^6}{6!}+O\left(x^7\right)\tag{2}
\end{align}
$$
The coefficient of $\frac{x^5}{5!}$ in $(2)$ is $2111$.
Best Answer
As pointed out in a comment, there are numerous errors in your answer.
Pres-Veep . . . . . Secretaries
W-W . . . . . . . . . 2W,1M or 1W-2M: $\left[{4\choose1 }{3\choose 1}\right]\left[{2\choose2}{4\choose1} + {2\choose 1}{4\choose 2}\right]= 192$
W-M or M-W . . . . 3W or 2W,1M: $\left[2{4\choose1}{4\choose1}\right]\left[{3\choose 3}+{3\choose2}{3\choose1}\right]=320$