Suppose an airline requires 3 pilots and 6 flight attendants to fly.
It has 12 pilots (9 men, 3 women) and 32 flight attendants (6 men, 26 women). The likelihood of choosing a specific person is equal.
What is the probability that a flight has exactly 2 men and 7 women?
My logic is to break this down into 3 cases: 0 male pilots/2 male flight attendants, 1 male pilot/1 male flight attendant, and 2 male pilots/0 male flight attendants. Then the overall probability will be the sum of these.
So for 0 pilots/2 FA: $$\frac {\binom 9 0 \binom 3 3 \binom 6 2 \binom {26} 4}{\binom {12} 3 \binom {32} 6 }$$
Repeating for all cases and adding them, I get:
$$\frac {\binom 9 0 \binom 3 3 \binom 6 2 \binom {26} 4}{\binom {12} 3 \binom {32} 6 }+\frac {\binom 9 1 \binom 3 2 \binom 6 1 \binom {26} 5}{\binom {12} 3 \binom {32} 6 }+\frac {\binom 9 2 \binom 3 1 \binom 6 0 \binom {26} 6}{\binom {12} 3 \binom {32} 6 }$$
But apparently this is wrong. Where is my logic flawed?
(Also sorry about poor formatting)
Your logic is not flawed at all, but your calculation seems to be. I find the numbers work out to approximately $0.1793$ (see here).