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Counting permutations uses combinatorics and factorials

For n items, the number of permutations is equal to n! (factorial of n)

How to count distinct permutations?

Having a repeated item involves a division of the number of permutations. One has to count the number of permutations of these repeated items.

DCODE has 5!=120 permutations but has 2 times D (2! permutations) so one divides 5! by 2! then 5!/2!=60

How to remove the limit when computing permutations?

Permutations makes exponential values witch needs huge computing servers, so the generation must be paid.

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