$\text{Method 1:}$
$$\left(\begin{array}{l}8 \\6\end{array}\right)=28$$
Then let's take away the choices where we don't cover all $4$ groups. There are only $4.$ Why? To not cover all $4$ groups but to choose $6$ means to pick all questions but $2$ questions of the same group. That is, from $4$ groups, pick $1$ group to exclude or equivalently pick $3$ groups to include
$$\begin{gathered} \left(\begin{array}{l} 4 \\3 \end{array}\right)=\left(\begin{array}{l} 4 \\ 1 \end{array}\right)=4 \\ \therefore\left(\begin{array}{l} 8 \\ 6 \end{array}\right)-\left(\begin{array}{l} 4 \\ 3 \end{array}\right)=\left(\begin{array}{l} 8 \\ 6
\end{array}\right)-\left(\begin{array}{l} 4 \\ 1 \end{array}\right)=28-4=24 \end{gathered} $$
$\text{Method 2:}$
Because we only have four groups and you need to pick up 6 balls which means you need to pick up two groups of balls and one ball each from the other two groups so the final result would be $\left(^{4} \mathrm{C} _{2}\right) \times 2 \times 2$ which is $24.$
$\text{Method 3:}$
The student can choose two questions to omit, not both in 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 chosen doesn't matter, so we get $8 \times 6 / 2=24$ options.
Alternatively, the student can choose two groups to omit a question from, and then one question from each group. This way we get $\left(^{4} \mathrm{C}_{2}\right)\left(^{2} \mathrm{C} _{1}\right)\left(^{2} \mathrm{C} _{1}\right)=6 \times 2 \times 2=24$ options.
