# Thread: derangements and permutations in cryptography

1. ## derangements and permutations in cryptography

hi, i have a problem that i am having a bit of trouble with;

we are given a partial key (missing 11 letters) for a mono-alphabetic substitution cipher and asked to calculate the number of possible keys given that no plaintext letter can be mapped to itself.

ordinarily, the number of possible keys would be the number of derangements of the missing letters (!11), however 5 of the plaintext letters that are missing mappings already exist as mappings in the partial key, so logically it shouldnt matter what the mapping of those plaintext letters is, because they can never map to themselves.

so shouldnt the number of possible keys be 5! * !6, ie. (the number of permutations of the 5 already mapped free letters) * (the number of derangements of the remaining 6)?

the problem is that 5! * !6 = 31800 which is much less than !11 = 39916800

intuitively the set of derangements should be a smaller subset of !11, shouldnt it?

am i just getting something wrong in my arithmetic? or am i completely missing the concepts? any help would be greatly appreciated

thanks gus

2. ## Re: derangements and permutations in cryptography

sorry, the value for !11 should be 14684570, not 39916800 which is the value of 11!..