Finding the number of councils with at least two girls There are 12 people, 7 girls and 5 boys.
You are supposed to choose a council of 4 people so that there are at least 2 girls.
I know you can solve this by taking the total amount of choices and subtracting the scenarios with one or no girls.
12C4 - (5C4) - (7C1)(5C3) = 420
But I'm pretty sure you can obtain the answer through other considerations.
If you assume that the first two seats must be girls, then the final two spots would have 10 and 9 people to choose from. So I reasoned that the answer must be 
7*6*10*9 = (7C2)(10C2) = 420*9
So I'm a factor of 9 off. My hunch is that my answer is double counting somehow but I'm having trouble identifying my mistake. Help?
 A: As lulu suggests in the comments, a way to count the number of councils with at least two girls is to add the number of ways of selecting a council with exactly two, exactly three, or exactly four girls.  
The number of ways of selecting exactly $k$ girls and $4 - k$ boys from  seven girls and five boys is $$\binom{7}{k}\binom{5}{4 - k}$$  Hence, the number of ways of selecting a council with at least two girls is $$\binom{7}{2}\binom{5}{2} + \binom{7}{3}\binom{5}{1} + \binom{7}{4}\binom{5}{0} = 21 \cdot 10 + 35 \cdot 5 + 35 \cdot 1 = 210 + 175 + 35 = 420$$ as you found by subtracting the number of cases in which fewer than two girls are selected from the number of ways of selecting four of the twelve available people.
Where did you go wrong?

If you assume that the first two seats must be girls, then the final two spots would have $10$ and $9$ people to choose from. So I reasoned that the answer must be $$7 \cdot 6 \cdot 10 \cdot 9 = \binom{7}{2}\binom{10}{2}$$

What matters here is who is selected, not the order in which they are selected.  Your count $7 \cdot 6 \cdot 10 \cdot 9$ is the number of ways of selecting a girl, a second girl, selecting one of the ten remaining people, and then selecting one of the remaining nine people in that order.  
Since the number of ways of selecting a subset of $k$ objects from a set of $n$ objects is 
$$\binom{n}{k} = \frac{n!}{k!(n - k)!}$$
the number of ways of selecting two objects from a set with $n$ objects is 
$$\binom{n}{2} = \frac{n!}{2!(n - 2)!} = \frac{n(n - 1)(n - 2)!}{2 \cdot 1 \cdot (n - 2)!} = \frac{n(n - 1)}{2}$$
In particular,
$$\binom{7}{2} = \frac{7 \cdot 6}{2} = 6 \cdot 3 = 21$$
and 
$$\binom{10}{2} = \frac{10 \cdot 9}{2} = 5 \cdot 9 = 45$$
Hence, the statement 
$$7 \cdot 6 \cdot 10 \cdot 9 = \binom{7}{2}\binom{10}{2}$$
is false since 
$$7 \cdot 6 \cdot 10 \cdot 9 = 42 \cdot 90 > 21 \cdot 45 = \binom{7}{2}\binom{10}{2}$$
The answer 
$$\binom{7}{2}\binom{10}{2}$$
is still too large.  To see why, suppose Anne, Belinda, Claire, and Edward are selected for the council.  You count this selection three times, once for each of the $\binom{3}{2}$ ways of designating two of the three women as the designated women.
$$
\begin{array}{l l}
\text{designated women} & \text{additional people}\\ \hline
\text{Anne, Belinda} & \text{Claire, Edward}\\
\text{Anne, Claire} & \text{Belinda, Edward}\\
\text{Belinda, Claire} & \text{Anne, Edward}
\end{array}
$$
Suppose Anne, Belinda, Claire, and Debra are selected for the council.  You count this selection six times, once for each of the $\binom{4}{2}$ ways of designating two of the four women as the designated women.
$$
\begin{array}{l l}
\text{designated women} & \text{additional people}\\ \hline
\text{Anne, Belinda} & \text{Claire, Debra}\\
\text{Anne, Claire} & \text{Belinda, Debra}\\
\text{Anne, Debra} & \text{Belinda, Claire}\\
\text{Belinda, Claire} & \text{Anne, Debra}\\
\text{Belinda, Debra} & \text{Anne, Claire}\\
\text{Claire, Debra} & \text{Anne, Belinda}
\end{array}
$$ 
Notice that 
$$\binom{2}{2}\binom{7}{2}\binom{5}{2} + \color{red}{\binom{3}{2}}\binom{7}{3}\binom{5}{1} + \color{red}{\binom{4}{2}}\binom{7}{4}\binom{5}{0} = \binom{7}{2}\binom{10}{2}$$
The reason you got too large an answer is that the group of seven girls and the group of ten additional people are not disjoint.
