Argh, i'm damn slow.. i was sure the answer was in the Tartaglia's triangle, but no way.

here my, plain as usual, solution:

use strict;
use warnings;

my @posts = (1..$ARGV[0] || 5);
my @pigeons = (1..$ARGV[1] || 3);
print +('1 ' x @pigeons)."\n" if @posts == @pigeons;
die  sprintf "%s is not enought for %s pigeons!",scalar @posts,scalar 
+@pigeons
     if @pigeons > @posts;

for (@pigeons) {
    my $max = @posts - @pigeons + 1;
    while ($max > 1){
          my @distr = (0) x ($#pigeons+1);
          $distr[0] = $max;
          my $remain = @posts - $max;
          my $i=0;
          while ($remain > 0){
                $i == $#distr ? $i=1 : $i++;
                $distr[$i]++;
                $remain--;
          }
          $max--;
          print +(map{ "$distr[$_-1] "} @pigeons),"\n";
    }
    @pigeons = (pop @pigeons,@pigeons);
}


__DATA__
perl BUk-pigeons.pl 5 3

3 1 1
2 2 1
1 3 1
1 2 2
1 1 3
2 1 2
[download]

thanks for the amusement!
At the end of writing i was surpised i needed not an hash to get uniques results..

L*

UPDATE: sadly my solutions gives too few combinations.. perl BUk-pigeons.pl 12 7 | wc -l give me only 35 combinations..