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,
$\binom {12}{2} \binom {10}{3} + \binom {12}{3} \binom {10}{2} + \binom {12}{4} \binom {10}{1} + \binom {12}{5}\binom {10}{0} = 23562$
Best Answer
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.