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Question Number 43344 by peter frank last updated on 10/Sep/18
The number 1, 2, 3, ..., n  are arranged  in a random order. The probability  that the digits 1, 2, 3, ..., k (k>n) appears  as neighbours is
$$\mathrm{The}\:\mathrm{number}\:\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:…,\:{n}\:\:\mathrm{are}\:\mathrm{arranged} \\ $$$$\mathrm{in}\:\mathrm{a}\:\mathrm{random}\:\mathrm{order}.\:\mathrm{The}\:\mathrm{probability} \\ $$$$\mathrm{that}\:\mathrm{the}\:\mathrm{digits}\:\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:…,\:{k}\:\left({k}>{n}\right)\:\mathrm{appears} \\ $$$$\mathrm{as}\:\mathrm{neighbours}\:\mathrm{is} \\ $$
Commented by MrW3 last updated on 10/Sep/18
k<n  (((n−k+1)!k!)/(n!))
$${k}<{n} \\ $$$$\frac{\left({n}−{k}+\mathrm{1}\right)!{k}!}{{n}!} \\ $$

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