Choosing three pairs out of eight items

3 students each choose two problems from a list of eight problems. How many ways can this be done?

The answer in the text book gives 8!/(2! 2! 2! 2!), but don't we also have to multiply by the number of ways the questions can be assigned to each student? So 3! x 8!/(2! 2! 2! 2!)?

We assume that Student 1 picks $2$ problems, then Student 2 picks $2$ problems from the remaining $6$, then Student $3$ picks $2$ problems from the remaining $4$.

Then S1 has $\binom{8}{2}$ choices, and for every such choice S2 has $\binom{6}{2}$ choices, and then S3 has $\binom{4}{2}$ choices, for a total of $\binom{8}{2}\binom{6}{2}\binom{4}{2}$,

This simplifies to the expression of your answer. Which student gets which problems is built into the calculation, we do not multiply by $3!$.

No. How does the $8!/(2!)^4$ arise?

1. First, we order the questions in a random way (that's $8!$).
2. Then give first two questions to the first student (disregarding the order between them, so we divide by $2!$).
3. Then we give the next two to the second, third pair to the third, and then there are two left. In each pair the order doesn't matter, so we divide by $(2!)^4$ in total.

Now, notice that we already are accounting for the ordering between the students: we're giving the first pair of questions to the first student, not the second pair (and so on).

the main point is that if you sepcify three students with $A$,$B$,$C$, which means that we distinguish these three student, then we can conclude the answer is $\{\begin{pmatrix} 3\\ 1 \end{pmatrix}+\begin{pmatrix} 2\\ 1 \end{pmatrix}+\begin{pmatrix} 1\\ 1 \end{pmatrix} \} \begin{pmatrix} 8\\ 2 \end{pmatrix}\begin{pmatrix} 6\\ 2 \end{pmatrix}\begin{pmatrix} 4\\ 2 \end{pmatrix}$, not $3!(\ldots)$