The word "combination" in the title brought Algorithm::Combinatorics to mind. I'm not sure if duplicate sums, e.g. the several 5+5+10+80 combinations in the third example, should all be shown but I have eliminated them. This code

use strict;
use warnings;

use feature qw{ say };

use Algorithm::Combinatorics qw{ combinations };
use List::Util               qw{ sum };

my @tests = (
   {
      target => 100,
      values => [ 1, 99, 2, 40, 50, 100, 60, 90, 3, 5, 95, 100 ],
   },
   {
      target => 10,
      values => [ 1, 3, 2, 4 ],
   },
   {
      target => 100,
      values => [ 5, 5, 5, 5, 10, 15, 80, 99 ],
   },
   );

foreach my $rhTest ( @tests )
{
    say
       qq{\nFind sums from },
       join( q{, }, @{ $rhTest->{ values } } ),
       qq{ making $rhTest->{ target }};

    say for do {
       my %seen;
       grep { ! $seen{ $_ } ++ }
       grep { $_ == $rhTest->{ target } }
       @{ $rhTest->{ values } };
       };

    for my $sumsOf ( 2 .. scalar @{ $rhTest->{ values } } )
    {
        my $combIter = combinations( $rhTest->{ values }, $sumsOf );
        my %seen;

        while ( my $raComb = $combIter->next() )
        {
            next if $seen{ join q{+}, sort { $a <=> $b } @{ $raComb } 
+} ++;
            say join q{+}, @{ $raComb }
               if $rhTest->{ target } == sum @{ $raComb };
        }
    }
}
[download]

produces

Find sums from 1, 99, 2, 40, 50, 100, 60, 90, 3, 5, 95, 100 making 100
100
1+99
40+60
5+95
2+3+95
2+90+3+5
2+40+50+3+5

Find sums from 1, 3, 2, 4 making 10
1+3+2+4

Find sums from 5, 5, 5, 5, 10, 15, 80, 99 making 100
5+15+80
5+5+10+80
5+5+5+5+80
[download]

I hope this is helpful.