Let's do this question by different approach:
Total permutations = n!
Part A: n precedes 1 and n −1 precedes 2
By symmetery Event 1: half of n! permutations have n preceding 1 and other half has 1 preceding n ,so probability = 1/2
Similarly Event 2: half of n! permutations have n-1 preceding 2 and other half has 2 preceding n-1 ,so probability = 1/2
Both events occur together so probabilty = 1/4 (as both events are independent of each other)
Part B: n precedes 1 and 2
Again By symmetery in total of n! permutations each of n or 1 or 2 can precede the other two(if n precede 1 and 2 then we don't care about how 1 and 2 are placed. We only care n precede both 1 and 2) therefore n precede both 1 and 2 in exactly($\frac{n!}{3}$) permutations so probability = 1/3