Looping on ranges (Modular iterations)

Question

Is it possible to make the following code modular, so that it works up till depth n, and returns an array of all possible combinations for used{}?

ranges is an nx2 array which contains the ranges for the ith loop.

    used = Table[0, lb]
    For[i = ranges[[1, 1]], i <= ranges[[1, 2]], i++,
     used[[1]] = i;
     For[j = ranges[[2, 1]], j <= ranges[[2, 2]], j++,
      If[FreeQ[used, j],
       used[[2]] = j;
       For[k = ranges[[3, 1]], k <= ranges[[3, 2]], k++,
        If[FreeQ[used, k],
         used[[3]] = k;
         For[l = ranges[[4, 1]], l <= ranges[[4, 2]], l++,
          If[FreeQ[used, l],
           used[[4]] = l;
           For[m = ranges[[5, 1]], m <= ranges[[5, 2]], m++,
            If[FreeQ[used, m],
             used[[5]] = m;
             (*operations*)

             used[[5]] = 0;
                  ];
                ];
              used[[4]] = 0;
              ];
            ];
          used[[3]] = 0;
          ];
        ];
      used[[2]] = 0;
      ];
    ];
  used[[1]] = 0
  ];

I am not able to find the proper tags, so suggestions are welcome.

Answer 1

Maybe the following does what you look for.

step = {x, ran} \[Function] Select[
    Flatten[Outer[Join, x, Partition[Range @@ ran, 1], 1], 1],
    DuplicateFreeQ
    ];

lb = 3;
ranges = Sort /@ RandomInteger[{1, 10}, {lb, 2}];
Join @@ FoldList[
  step,
  {{}},
  ranges
  ]

Maybe you are only interested in (the result of) Select[Tuples[Range @@@ ranges],DuplicateFreeQ]. Then

Fold[step, {{}}, ranges]

provides a much faster alternative (compare for lb = 10).