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What is the probability that a random permutation of the integers $\{1,\dots,n\}$ has no two consecutive numbers next to each other?

For example, for $n=4$ the permutation $2,4,1,3$ has no two consecutive numbers next to each other. However, $1,2,3,4$ and $4,3,2,1$ do. For $n = 3$ the probability is $0$.

  • If $n=4$ the probability is $1/12$.
  • If $n=5$ the probability is $7/60$.
  • If $n=6$ the probability is $1/8$.
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    The number of permutations of $1,2,3,\dots,n$ with no consecutive numbers adjacent is given by OEIS A002464. Divide by $n!$ to get the probability. https://oeis.org/A002464 – awkward May 15 '18 at 15:17
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    https://math.stackexchange.com/questions/1822068/how-many-permutations-of-1-2-3-n-there-are-with-no-2-consecutive-numbe – user614287 Apr 30 '19 at 22:38

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