Respuesta :
We may count the total number of letter permutations and then deduct the number of permutations in which at least one letter is present in its original location to get the number of permutations in which no letter appears in its original place.
A collection of n letters can be arranged in n! different ways, where n! is the factorial of n, which is the sum of the first n positive numbers. The set "a, b, c," for instance, has the following 3! = 6 permutations: (a, b, c), (a, c, b), (b, a, c), (b, c, a), (c, a, b), and (c, b, a).
|A[1] A[2] ... A[n]| equals |A[1]| + |A[2]| +... + |A[n]| minus |A[1] A[2]| minus |A[1] A[3]| minus... - |A[n-1] A[n]| + |A[1] A[2] A[3]|
The first term adds the sizes of each set, the second word adds the sizes of their intersections, and so on.
Therefore, there are 3! - 3 = 3 permutations in which no letter appears in its original position.
Learn more about the permutation and combination at
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