Number of committees of size $5$ with at least $2$ women from a society with $10$ men and $12$ women I've been thinking about this problem:
A committee of size $5$ is formed from a society with a membership of $10$ men and $12$ women, with the restriction that there are at least $2$ women on the committee. How many committees are there?
I would like help understanding the flaws in my first approach:
Argue that every such committee could be formed by first choosing $2$ women  to be on the committee and then filling the remaining $3$ places.
There are $\binom {12}{2}$ and $\binom {20}{3}$ ways to achieve each task and so the total number of committees  is $\binom {12}{2} \binom {20}{3}$.
However this a dramatic over count since  $\binom {12}{2} \binom {20}{3} = 75240 > 26334 =  \binom {22}{5}$.
Although I know this answer is false, I am struggling to pinpoint my faulty reasoning. Where are the mistakes and can the argument be salvaged?

Note:
  I am aware that the commitees can be partitioned according to the number of women in the committee to yield the answer,
  $\binom {12}{2} \binom {10}{3} +  \binom {12}{3} \binom {10}{2} + \binom {12}{4} \binom {10}{1} + \binom {12}{5}\binom {10}{0} = 23562$

 A: The fault is that, if a committee with more than two women is chosen, then the two women you pre-selected play a special role, so to say. The extra information involved in tagging them causes the number to be too large.
For example, you count each of the possible women-only committees ${5\choose 2}=10$-fold.
By the way, an alternative (also correct)  approach would be to first count the $22\choose 5$ committees we'd have without gender preference; then subtract the ${12\choose 0}{10\choose 5}$ and  ${12\choose 1}{10\choose 4}$ possibilities that have less than two women.
A: I think, because, the domains for which you calculated each ways are not totally disjoint. First you did C(12, 2), which is the number of ways 2 women can be chosen from 12 women. It feels like since 2 are chosen, remaining people are 20 (in fact it is so) but while calculating the ways of selecting 3 from the remaining 20, you cannot simply do C(20, 3) because 10 out of 20 would have already contributed for the combination in the earlier C(12, 2).
Just think of it like you do in probability, we can only multiply the probabilities only if they are mutually independent. Same is the case here, two domains are not independent.
