8. show that the probability that all permutations of the sequence 1,2,…,n have no number being still in the ith position is less than 0, 37 if n is large enough.
This is question on dearrangement
In combinatorial mathematics, a derangement is a permutation of the elements of a set, such that no element appears in its original position. In other words, a derangement is a permutation that has no fixed points.
the number of permutation of n distinct object is n!
hence required probability is Dn/n!
Now Dn = n!/e as n tend to infinity
hence
required probability = 1/e = 0.3678794411 < 0.37
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