Recursive permutation - recursion

Question

So I'm trying to permute all possible n digit long numbers out of x long array/set of elements. I've come up with a code that does that, however the digits are the same, how do I prevent that from happening. Here's my come(Pascal):
program Noname10;
var stop : boolean;
A : array[1..100] of integer;
function check( n : integer ) : boolean;
begin
if n = 343 // sets the limit when to stop.
then check := true
else check := false;
end;
procedure permute(p,result : integer);
var i : integer;
begin
if not stop
then if p = 0 then
begin
WriteLn(result);
if check(result)
then stop := true
end
else for i := 1 to 9 do
begin
permute(p - 1, 10*result+i);
end;
end;
begin
stop := false;
permute(3,0);
readln;
end.

Answer 1

Here is the code in Prolog
permutate(As,[B|Cs]) :- select(B, As, Bs), permutate(Bs, Cs).
select(A, [A|As], As).
select(A, [B|Bs], [B|Cs]) :- select(A, Bs, Cs).
?- permutate([a,b,c], P).
Pascal is much harder.
Here is an usefull algorithm, you might want to use. But it is not tested, so you have to debug it yourself. So you have to know how the algorithm works.
The Bell Permutation algorithm: http://programminggeeks.com/bell-algorithm-for-permutation/
procedure permutate(var numbers: array [1..100] of integer; size: integer;
var pos, dir: integer)
begin
if pos >= size then
begin
dir = -1 * dir;
swap(numbers, 1, 2);
end
else if pos < 1 then
begin
dir = -1 * dir;
swap(numbers, size-1, size);
end
else
begin
swap(numbers, pos, pos+1);
end;
pos = pos + dir;
end;
begin
var a, b: integer;
a = 1; b = 1;
while true do
begin
permutate(A, 5, a, b);
printArray(A, 5);
end;
end.