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What is the probability of these events when we randomly select a permutation of {1, 2, . . . , n} where n ≥ 4?

a) n precedes 1 and n −1 precedes 2.
b) n precedes 1 and n precedes 2.

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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

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