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...