Line up items in a row on a table. Now shuffle the items around and then re-arrange them in a row again. Let be the probability that there is no item in the same position as it was before it was shuffled. That is, all items were moved to a different position (for example, this would occur if all items were cycled one position to the right). Then,
for positive integers . Submit .
Note: We assume that, after shuffling, every permutation of the items is equally likely.
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The number of arrangements of n elements is n ! . The number of derangements is ! n = n ! k = 0 ∑ n k ! ( − 1 ) k ≈ n ! k = 0 ∑ ∞ k ! ( − 1 ) k = n ! e − 1 . The probability that an arrangement is in fact a derangement is n ! ! n ≈ n ! n ! e − 1 = e 1 1 . This approximation improves as n increases. It is clear that a = b = 1 , so the answer is 1 + 1 = 2 .