This problem can be solved using the method of cases, as suggested in the hint. The strategy involves breaking down the possibilities based on the behavior of two special groups: the four students who insist on being together (the "friends"), and the two students who won't work together (the "enemies").

We’ll look at different cases for the friends and enemies and compute how many ways we can choose the remaining students in each case.

Case 1: All 4 friends are chosen

In this case, we already have 4 students selected (the 4 friends), so we need to choose 6 more from the remaining 21 students. However, we must account for the constraint that we can select either 0 or 1 of the enemies, but not both.

Sub-case 1.1: Choose neither enemy

If we select none of the enemies, we must choose 6 regular students from the 19 remaining students. This can be done in: $\binom{19}{6} \text{ ways.}$

Sub-case 1.2: Choose exactly one enemy

If we select exactly 1 enemy, we must choose 5 more regular students from the remaining 19 students. Since there are 2 enemies, we can select one of them in $\binom{2}{1} = 2$ ways, and then choose 5 more students from the remaining 19 students. This can be done in: $2 \times \binom{19}{5} \text{ ways.}$

So, the total number of ways in Case 1 is: [ \binom{19}{6} + 2 \