A teacher divides seven girls and five boys randomly into three groups of four each. What is the probability that one group has four girls? Question in title. Sorry, this is my first question and I don't know how to format.
My thinking was 
$$\frac{3\binom{7}{4}\binom{8}{4}\binom{4}{4}}{\binom{12}{4}\binom{8}{4}\binom{4}{4}}.$$
 The numerator comes from 3 groups to put four girls into, and $\binom{7}{4}$ ways to choose those girls. Once that group is set, there are $\binom{8}{4}$ to put kids into the second group and $\binom{4}{4}$ ways to put kids into the third group.
That gives me an answer of $7/132$ but the book I'm using says the answer is $7/33$. Can you please help me find what I'm missing? Thank you!

Your method of solution is indeed correct.  Check your arithmetic.
$$\frac{3\cdot \binom{7}{4} \binom{8}{4} \binom{4}{4}}{\binom{12}{4}\binom{8}{4}\binom{4}{4}} = \frac{3\cdot \binom{7}{4}}{\binom{12}{4}}=\frac{3\cdot 35}{495}=\frac{7}{33}$$
