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Let $\pi$ be a permutation of the integers $0$ through $2^n-1$ (the integers that can be represented in $n$ bits). For example, $\pi$ could be used as a substitution cipher in which each plaintext block is $n$ bits long. How many possible choices of permutation $\pi$ are there? Equivalently, how many permutations are there of the first $2^n$ non-negative integers?

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Please do some research before asking obvious questions, and doing that poorly: there is a typo (the first $2^n$ non-negative integers is obviously met). Also, π defined in the first sentence is not used in the second, which forms an independent question; so why include the definition of π? –  fgrieu Jan 28 at 7:14
    
@fgrieu: To be fair, I don't see a typo, the question shows $[0,2^n-1]$? Its still a low quality question, at best containing combinatorics taught in secondary school (UK). –  figlesquidge Jan 28 at 10:40
    
Typo in the question is fixed thanks to figlesquidge; typo in my comment remains (met should be meant). –  fgrieu Jan 29 at 14:28
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closed as off-topic by fgrieu, DrLecter, rath, figlesquidge, Maeher Jan 28 at 10:35

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1 Answer

There are $k!=k(k-1)\dots3\cdot 2\cdot 1$ possible permutations of $k$ elements.

This is very basic combinatorics.

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