G'day SaktiMonk,

Welcome to the monastery.

Update: My original solution (now within <spoiler> tags below) didn't take into account uneven sized ranges. I didn't notice that but ++Laurent_R did (see below). Here's a new version that still uses an array (@bins) for the range counts; caches the results of determining the range; and still only uses one iteration per range to output the bin counts.

$ perl -Mstrict -Mwarnings -e '
    use List::Util qw{first};

    my @input = qw{12 14 15 20 21};
    my @range_ends = qw{19 29 39};
    my (@bins, %range_index_cache);

    sub get_range_index {
        ( first { $_[0] <= $range_ends[$_] } 0 .. $#range_ends ) // @r
+ange_ends;
    }

    for (@input) {
        ++$bins[$range_index_cache{$_} //= get_range_index($_)];
    }

    for (0 .. $#range_ends) {
        printf "%d-%d %d\n" =>
            $_ ? $range_ends[$_ - 1] + 1 : 1, $range_ends[$_], $bins[$
+_] // 0;
    }
'
1-19 3
20-29 2
30-39 0
[download]

Any datapoints beyond the ranges you specify are counted in the last element of @bins.

$ perl -Mstrict -Mwarnings -e '
    my @input = qw{12 14 15 20 21};
    my $ranges = 3;
    my @bins;

    for (@input) {
        ++$bins[$_ / 20];
    }

    for (0 .. $ranges - 1) {
        printf "%d-%d %d\n" => $_ * 20 || 1, $_ * 20 + 19, $bins[$_] /
+/ 0;
    }
'
1-19 3
20-39 2
40-59 0
[download]

[Note: If you're using an old version of Perl that doesn't have the // (Logical Defined-Or) operator, you'll need to use the defined function.]