In how many ways can a $31$ member management be selected from $40$ men and $40$ women so that there is a majority of women? Here's the question:

In an organization there are $80$ people, $40$ men and $40$ women.
  In how many ways can we choose, from those $80$ people, a $31$ member management so that there is a majority of women? 

My answer was to first choose $16$ women (to assure the majority) out of the $40$ meaning $\binom{40}{16}$,
and then choose the rest $15$ from the rest of all people meaning $\binom{64}{15}$. So the final answer is ${40 \choose 16} \cdot{64 \choose 15}$ but that isn't the right answer so I really don't know..
help?
Thank you.

Note $\binom{n}{k} = \frac{n!}{k!(n-k)!}$, the binomial coefficient.
 A: The reason your solution gave the wrong answer is that you are sometimes counting the same committee more than once, e.g. if you:


*

*choose women #1..#16 and then choose woman #17 + men #1..14


that's the same as


*

*choose women #1..#15,woman#17 and then choose woman #16 + men #1..14


One productive way of thinking about the question is to ask "how many ways are there of choosing a committee with an equal number of men and women?", then think about the other committees: What proportion of them have more men than women? What proportion have more women than men?
Edit: that's not such a productive line of thought. See DJohnM's answer for where to go next.
A: The correct calculation using your logic should go as follows:
Remember we are looking for the number of committees where there is a majority of women.
Well, it could be the case that there 16 women and 15 men, and this should be
$$\binom{40}{16}\binom{40}{15}.$$
But this is not the only case; it could be the case that there are 17 women and 14 men:
$$\binom{40}{17}\binom{40}{14}.$$
Notice that since the composition/ratio of men and women are different, we are not counting the same committees.
Following this logic, we add up all the cases which is
$$\sum_{k = 16}^{31}\binom{40}{k}\binom{40}{31-k}.$$
But this is not the ideal method/answer. You want to approach it as the other answer/post suggests.
A: How many 31 member committees can you make while ignoring the gender mix?
That would be $\binom{80}{31}$
What fraction of those will have a female majority? Think symmetry, since you have a lot of it in this case.
