I was just looking at the code here and it struck me that there must be a simpler way than having to mess with lists.

Suppose I have to pick just one random number from the range 1..10. I can do it by using the following code:

sub pick_one
  {
    my($from,$to) = @_;
    for(my $i=$from;$i<=$to;$i++)
      {
        return $i if(rand(1) <= 1/($to-$i+1));
      }
  }
[download]

(Notice that when $i==$to the test must always succeed so we don't have to worry about a final return)

If I have to pick $n more numbers and I have ($to-$from+1) numbers left then the chance that the next number will be one of those picked is ($n/($to-$from+1)), so the final form of the subroutine is:

sub pick_n
  {
    my($n,$from,$to) = @_;
    my(@picked);

    for(my $i=$from;$i<=$to;$i++)
      {
        if(rand(1) <= ($n/($to-$i+1)))
          {
            push(@picked,$i);
            $n--;
          }
      }
    return @picked;
  }
[download]

Without having to shuffle lists or even use them. Notice the time is always the proportional to the range and the results are always unique. Just to prove that the results are fair:

my(@counts);

for(my $i=0;$i<100000;$i++)
  {
    foreach my $c (pick_n(3,1,10))
      {
        $counts[$c]++;
      }
  }

for(my $c=0;$c<=$#counts;$c++)
  {
    print "Count[$c] = $counts[$c]\n";
  }
[download]

=>

Count[0] =
Count[1] = 29974
Count[2] = 30044
Count[3] = 29860
Count[4] = 29949
Count[5] = 30167
Count[6] = 30253
Count[7] = 29808
Count[8] = 30092
Count[9] = 29830
Count[10] = 30028
[download]

Update: This algorithm will work well if there are lots of numbers to be selected from a shorter list. If selecting fewer numbers from longer lists then the simple hash approach that most other suggestions are based on will be more efficient.