TITLE: Question on Permutation/Combination
QUESTION [5 upvotes]: A committee of 12 is to be formed from 9 women and 8 men. In how many ways this can be done if at least 5 women have to be included in a committee?
I don't want to know how to answer this. I know that we have to take cases (5 women, 6 women, 7 women and 8 women) and add them up. 
I want to know why I'm not getting the same result if I take 5 women first and then merge the women and men and take 7 people from them.
This will give me ${9\choose5}\times{12\choose7}$ but the answer given is 6062, which I do get by taking cases.

REPLY [3 votes]: Others have explained what is wrong with your calculation. Here is a shorter right approach than what you sketch:
There are $\binom{17}{12}$ possible committees if we ignore the gender quota.
The only way there can possibly be less than $5$ women on the committee is if all $8$ men are on it -- so $\binom{9}{4}$ of the $\binom{17}{12}$ possible committees are invalid.
The number of valid committees is therefore
$$ \binom{17}{12} - \binom{9}{4}. $$