Forming a committee from $4$ gentlemen and $4$ ladies with certain conditions 
From $4$ gentlemen and $4$ ladies a committee of $5$ is to be formed .
  If the committee consists of $1$ president, $1$ vice president and $3$ secretaries.
  What will be the number of ways of selecting the committee with at least $3$
  women such that at least one women holds the post of president or vice-president?

I tried 
Case $1$
$3W2M$
$\dbinom{4}{1}\times \dbinom{4}{1} \times \dbinom{3}{2} \times \dbinom{3}{1}=144 $
$+$
$\dbinom{4}{1}\times \dbinom{3}{1} \times \dbinom{2}{1} \times \dbinom{4}{2}=144 $
Case $2$
$4W1M$
$\dbinom{4}{1}\times \dbinom{3}{1} \times \dbinom{2}{2} \times \dbinom{3}{1}=36 $
$+$
$\dbinom{4}{1}\times \dbinom{4}{1} \times \dbinom{3}{3}=16 $
Total ways=$338$.
But the answer given is $512$.
I look for a short and simple way.
I have studied maths up to $12$th grade.

Count the number of committees with $\color\red{3\text{ women}}$ and $\color\green{2\text{ men}}$:
$$\color\red{\binom43}\cdot\color\green{\binom42}\cdot\binom51\cdot\binom41\cdot\binom33=480$$
Subtract the number of committees where all $3$ women are secretaries:
$$\color\red{\binom43}\cdot\color\green{\binom42}\cdot\color\green{\binom21}\cdot\color\green{\binom11}\cdot\color\red{\binom33}=48$$
Add the number of committees with $\color\red{4\text{ women}}$ and $\color\green{1\text{ man}}$:
$$\color\red{\binom44}\cdot\color\green{\binom41}\cdot\binom51\cdot\binom41\cdot\binom33=80$$
Hence the number of committees is $480-48+80=512$.
