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.

The student can choose two questions to omit, not both i  the same group. There are $8$ options for the first question, and then there are $6$ options left for the second question. Of course the order in which the questions are chosrn doesn't matter, so we get
$$\frac{8\times6}{2}=24$$
options.
Alternatively, the student can choose two groups to omit a question from, and then one question from each group. This way we get
$$\binom{4}{2}\binom{2}{1}\binom{2}{1}=6\times2\times2=24$$
options.
Finally, in line with your own approach, we can first choose one question from each group. We can indeed do so in $16$ ways. Then we can choose two more questions, indeed in $6$ ways. But now we have overcounted; we reach the same set of questions if we had first chosen other questions in the two groups we ended up choosing both questions from. In how many ways could we have chosen questions from these two groups? A total of $4$ ways. So by this method we have coumted each set of questions $4$ times. Hence the total number of options is $\frac{96}{4}=24$.
