Although there are only a few members of this club, the multiple conditions can make this problem seem intimidating.
Break it down into cases, and solve that way.
If Edward doesn’t go, then Paul doesn`t go. That leaves four students to fill four slots, so there is exactly one valid combination.
If Edward goes, then Milner won’t go for sure. The leaves four students left to take three slots, so 4C3 = 4.
That leaves us with a total of 1 + 4 = 5 valid combinations.
The correct answer is C.