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Jan 16, 2023 at 3:23 comment added David Pechersky So in the answer above (unlike my answer to your first question), I'm not intersecting with {1,2,3,...,k}. If we take the images of {1,2,3,...,k} under two random permutations and we also intersect this with {1,2,3,...,k}, this is the same (in distribution) as intersecting three random permutations. As per my answer above, this is concentrated around n*(k/n)^3, which matches what you got in your simulation.
Jan 15, 2023 at 10:13 comment added Alexander Chervov Thank you very much for your excellent answers ! The only thing - simulation suggests the leading coefficient is x^3/N^2 - when we have two permutations. E.g. kaggle.com/code/alexandervc/…
Jan 11, 2023 at 9:50 vote accept Alexander Chervov
Jan 9, 2023 at 20:46 history answered David Pechersky CC BY-SA 4.0