With 435 zeroes and 343 ones, you'll never live long enough to see it finish, regardless of the method used to generate the sequence. A simple way to figure out how many permutations a list will generate is with 
my $num_zeros = 3; my $num_ones = 2;
use Math::BigInt;
sub fac {
    my $result = Math::BigInt->new("1");
    my $n = Math::BigInt->new(shift());
    while($n){ $result *= $n--; }
    return $result;
}
my $num_permutations = fac($num_ones + $num_zeros ) / (fac($num_ones) 
+* fac($num_zeros));
print("Number of permutations of a ($num_zeros, $num_ones) list is $nu
+m_permutations\n");'

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A list with 453 zeros and 343 ones will have roughly 1.962e230 different permutations -- are you sure you need all of them?

In reply to Re^5: One Zero variants_without_repetition by ohcamacj
in thread One Zero variants_without_repetition by thenetfreaker

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