I like to match permutations of 0-8 with each digits occurring once using all 8 digits.

Um... the way I learned it, "0-8" represents 9 digits. (1-8 would be 8 digits, as would 0-7.) Is there some digit between 0 and 8 that you intend to leave out, and if so, which one?

Anyway, if you had said "permutations of 0-8 with each digit occurring once, using all 9 digits", then I would understand that you are looking only for strings of nine digit characters, such that all nine characters are distinct, and none of them is the digit "9":

#!/usr/bin/perl

use strict;

while (<DATA>) {
    chomp;
    if ( length() == 9 and not ( /[^0-8]/ or /(.).*\1/ )) {
        print "$_\n";
    } else {
        warn "rejected input: $_\n";
    }
}

__DATA__
01234
123456780
1234567890
223456781
345678012
234567890
a12345678
012345678
456782011
456782013
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Update: I suspect that the regex used as the last stage in my conditional is a fairly expensive operation; for strings that actually meet the criteria (are not rejected), it has to do 8+7+6+...+1 (total of 36) character comparisons to finish. (There should be some sort of "O(...n...)" expression for that, but it escapes me.) So, it would most likely be better to use a split/hash solution, as suggested by others, especially if you'll be handling large quantities of input with a relatively high "success" rate. Something like this:

while (<DATA>) {
    chomp;
    if ( length() == 9 and not /[^0-8]/ ) {
        my %c = map { $_ => undef } split //;
        if ( keys %c == 9 ) {
            print "$_\n";
            next;
        }
    }
    warn "rejected input: $_\n";
}
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