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I How to find the nth binary permutation?

  1. Jan 16, 2017 #1
    I need to complete this.
    Let's say I have 4 bits with 2 bits set as 1, 0011. The total number of permutations for this number is 0011, 0101, 0110, 1001, 1010, 1100, 6 cases. This can be computed using the calculation.

    4! / ((2!)(4-2)!) = 6

    Now I want to be able to find the nth sequence, for instance 1st number is 0011, second number is 0101. So if I say n=5, I want to be able to get the 5th permutation sequence 1010 from the initial 0011. How do I do this?
  2. jcsd
  3. Jan 16, 2017 #2


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    What do you mean by 'the nth sequence'? You have given a 'for instance' but I don't even understand that case, let alone how you wish to generalise it.
  4. Jan 16, 2017 #3


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    There are algorithms to find the nth permutation for a given order (e. g. lexicographic - as it would appear in a lexikon). Example, another example.

    The basic idea: Start with the first digit: How many permutations start with 0? Compare your number to that number and you get the first digit. Continue like that for all other digits.
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