GRE combinatorics question about counting the no. of sets questions satisfying a certain requirement. From ETS Major Field Test in Mathematics

A student is given an exam consisting of
  8 essay questions divided into 4 groups of
  2 questions each. The student is required to
  select a set of 6 questions to answer,
  including at least 1 question from each of
  the 4 groups. How many sets of questions
  satisfy this requirement? 

I'm thinking $$\binom{2}{1}^4 \binom{4}{2}$$
because we have to pick 1 from each of the 4 groups of 2 and then from the remaining 4 questions we pick 2.


I'm thinking $$\binom{2}{1}^4 \binom{4}{2}$$
because we have to pick 1 from each of the 4 groups of 2 and then from the remaining 4 questions we pick 2.

You are overcounting here: there will be two groups from which you end up picking both questions, and there is of course only one way to pick two questions from a group of two: pick both!
However, by first picking $1$ out of two (which can be done in two ways), and then picking the other one, your method ends up counting two ways to pick both questions from a group ... but those two ways are really just two different orderings of the group of two, and ordering is not a consideration for this question. Therefore, you overcount by a factor of $2$ for each of the groups where you end up picking both questions.
Finally, since there are two such groups, you overcount by a factor of $4$.
