Given a set of elements N = {1, 2, ..., n} and two arbitrary non-empty subsets…

Concept: for a permutation drawn uniformly from the n! permutations of N, symmetry says every element of a fixed subset S is equally likely to be the one whose image is smallest among π(S) -- so a single chosen element of S is that arg-min for exactly n!/|S| of the permutations.

For min(π(A)) = min(π(B)) to hold, the shared smallest image can only originate at an element that belongs to both A and B: an element outside A ∩ B lies in at most one of A, B, so it could be the minimum image of at most one of the two sets, never both.

1. If x ∈ A ∩ B is the element whose image π(x) is the smallest among π(A ∪ B), then since A, B ⊆ A ∪ B, π(x) is automatically the smallest of both π(A) and π(B), so min(π(A)) = min(π(B)) = π(x).

2. Conversely, whenever min(π(A)) = min(π(B)), that common value equals min(π(A ∪ B)) (because min(π(A ∪ B)) is the smaller of min(π(A)) and min(π(B))), and it is attained at an element of A ∩ B -- an element outside the intersection lies in only one of A, B and so cannot be the minimum image of both.

3. By the symmetry argument, for any single fixed element y ∈ A ∪ B, the number of permutations in which π(y) is the smallest image among π(A ∪ B) is n!/|A ∪ B|.

4. Summing this count over each of the |A ∩ B| elements that qualify gives the total count: |A ∩ B| × n!/|A ∪ B|.

Cross-check with n = 3, A = {1, 2}, B = {2, 3}: here |A ∩ B| = 1 and |A ∪ B| = 3, so the formula gives 3!·1/3 = 2. Listing all 6 permutations of {1, 2, 3} directly (written as (π(1), π(2), π(3))), exactly two of them -- (2, 1, 3) and (3, 1, 2) -- satisfy min(π(A)) = min(π(B)) = 1, confirming the count.