@Statistico Questions 24 and 25 are typically the most interesting ones on these papers. It's true that there hasn't been a combinatorics problem in those spots for several years, but that doesn't mean that this one was particularly difficult. I don't like their solution and think the problem is simpler than it implies.
First you need to recognize there are two cases: no ties and with one tie. Once you do that, it's obvious that there can be no overlap.
The case with no ties is straightforward enough: four ways for one position, three ways for another position, two ways for another position and one way for the remaining position. The product rule 4x3x2x1 = 24.
The case with ties splits into three sub-cases: 1st & 2nd are tied, 2nd & 3rd are tied, 3rd & 4th are tied. These sub-cases are equivalent in that there are two individual positions and one tied place, so we have to work out only one sub-case and can multiply by 3. There are four ways to choose one of the individual positions, three ways to choose the other individual position and... only one way to choose the remaining two people in the tie. So 4x3x1 = 12 for each sub-case, 3x12 = 36 for the case with ties.
However, it is true that contestants will be strong or weak in different areas, e.g., combinatorics, geometry, number theory, and algebra, so I will accept that my statement is probably too general.
As an aside, I think this year's Question 24 is rather easy for its slot. The only saving grace is that many children will have either forgotten to, or run out of time to, check the feasibility of N = 5 as a solution.