Question on Permutation/Combination 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. 
 A: The correct answer as you noted is:
$${9\choose5}{8\choose7}+{9\choose6}{8\choose6}+{9\choose7}{8\choose5}+{9\choose8}{8\choose4}+{9\choose9}{8\choose3}$$
Your alternate proposal is:
$${9\choose5}{12\choose7}$$
This is equal to:
$${9\choose5}{8\choose7}{5\choose5}+{9\choose6}{8\choose6}{6\choose5}+{9\choose7}{8\choose5}{7\choose5}+{9\choose8}{8\choose4}{8\choose5}+{9\choose9}{8\choose3}{9\choose5}$$
In other words, for each possible committee, you are over-counting that possibility by $n\choose5$ times, where $n$ is the number of women in that particular committee.  Do you see why?
A: You are overcounting, and let us see why.
Well, suppose your ladies are  named "a,b,c,d,e,f,g,h,i", and your men are named "j,k,l,m,n,o,p,q" (I apologize, indeed I named my fishes after letters of the Cyrillic script, so this does happen).
Suppose we choose 5 women first, let us say $a,b,c,d,e$. Then, later on you combine the rest and choose seven more people. Let us say I chose $f,g,j,k,l,m,n$, which is five men and two women, from the remaining bunch, to make a committee of twelve members, namely $a,b,c,d,e,f,g,j,k,l,m,n$.
Now, let's assume I went this way : I chose five women first, say $c,d,e,f,g$, and then I combined the rest of the people, and chose seven out of these, who turned out to be $a,b,j,k,l,m,n$. Then, this makes a committee of twelve members, which is $a,b,c,d,e,f,g,j,k,l,m,n$. 
The same committee has been formed in two "different" ways by your methodology. Hence, you are likely to be overcounting. This is reflected in your answer, which is $99,792$, a huge overestimate on the proposed answer of six thousand sixty two. 
A: Suppose the $9$ women's names are $A,B,C,D,E,F,G,H,I.$
And the $8$ men's names are $J,K,L,M,N,O,P,Q.$
At your first stage you pick $5$ women; suppose these are $A,B,C,D,E.$
At your second stage you pick $7$ others, who may be either women or men; suppose these are $F,G,J,K,L,M,N$ --- thus two women and five men.
But alternatively, suppose at your first stage the $5$ women you picked were $A,B,C,F,G.$
And then at your second stage you pick $7$ others and they are $D,E,J,K,L,M,N$ --- thus two women and five men.
Either way, you get the same set of seven women and five men.
But by your second method, you have counted this particular outcome twice (so far....). That's what's wrong with your second method.
A: 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
$$ \binom{17}{12} - \binom{9}{4}. $$
