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Given some natural number $N$, I am interested in the set of all binary permutations of length $N$ (with the intention of storing in lists depending on how many $1$'s appear in each permutation). My code does the job but is not efficient as I am using nested For loops. Please advise on how you would adapt for more efficient evaluation.

My code (for $N=3$):

Num = 3; A = {};
For[q = 0,  q <= Num, q++, B = {}; 
 For[i = 1, i <= Num - 1, i++, If[i <= q, AppendTo[B, 0]]; 
  If[i >=   q, AppendTo[B, 1]]]; 
 If[Length[B] < Num, AppendTo[B, B[[1]]]]; 
 AppendTo[A, Pe = Permutations[B]] ]

A = Reverse[A]

{{{0, 0, 0}}, {{0, 0, 1}, {0, 1, 0}, {1, 0, 0}}, {{0, 1, 1}, {1, 0, 
   1}, {1, 1, 0}}, {{1, 1, 1}}}

Thanks for any assistance.

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    $\begingroup$ In[1353]:= perms[n_] := SortBy[Map[IntegerDigits[#, 2, n] &, Range[0, 2^n - 1]], Total] In[1354]:= perms[3] Out[1354]= {{0, 0, 0}, {0, 0, 1}, {0, 1, 0}, {1, 0, 0}, {0, 1, 1}, {1, 0, 1}, {1, 1, 0}, {1, 1, 1}} ?? $\endgroup$ Oct 18 at 15:00
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GroupBy[Tuples[{0, 1}, 3], Total]
(*    <|0 -> {{0, 0, 0}},
        1 -> {{0, 0, 1}, {0, 1, 0}, {1, 0, 0}},
        2 -> {{0, 1, 1}, {1, 0, 1}, {1, 1, 0}},
        3 -> {{1, 1, 1}}|>                         *)

GatherBy[Tuples[{0, 1}, 3], Total]
(*    {{{0, 0, 0}},
       {{0, 0, 1}, {0, 1, 0}, {1, 0, 0}},
       {{0, 1, 1}, {1, 0, 1}, {1, 1, 0}},
       {{1, 1, 1}}}                          *)
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