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Here goes a larger (functional) code fragment, along with a test-case. The results of the test-case are below as well.

use strict; sub getDistinctPermutations($$); ### RUN SUB my @data = ('A', 'B', 'C', 'D', 'E', 'F'); my $cardinality = 2; my $dataref = getDistinctPermutations(\@data, $cardinality); if ($dataref == 0) { die "Bad return\n"; } for (my $i = 0; $i < scalar(@$dataref); $i++) { my $element = $$dataref[$i]; print join(' ', @{$element->{first}}, @{$element->{second}}), "\n" +; } ### SUB sub getDistinctPermutations($$) { my $permlist = $_[0]; my $cardinality = $_[1]; if (scalar(@$permlist) < 2) { return 0; } elsif ($cardinality < 1) { return 0; } my @data; if ($cardinality == 1) { for (my $i = 0; $i < scalar(@$permlist); $i++) { for (my $j = 0; $j < scalar(@$permlist); $j++) { push @data, { first => [$$permlist[$i]], second => [$$permlist[$j]], }; } } } if ($cardinality == 2) { for (my $i = 0; $i < scalar(@$permlist); $i++) { for (my $j = $i+1; $j < scalar(@$permlist); $j++) { for (my $k = 0; $k < scalar(@$permlist); $k++) { for (my $p = $k+1; $p < scalar(@$permlist); $p++) { push @data, { first => [$$permlist[$i], $$permlist[$j]], second => [$$permlist[$k], $$permlist[$p]], }; } } } } } return \@data; }

Which gives the results:

A B A B A B A C A B A D A B A E A B A F A B B C A B B D A B B E A B B F A B C D A B C E A B C F A B D E A B D F A B E F A C A B A C A C A C A D A C A E A C A F A C B C A C B D A C B E A C B F A C C D A C C E A C C F A C D E A C D F A C E F A D A B A D A C A D A D A D A E A D A F A D B C A D B D A D B E A D B F A D C D A D C E A D C F A D D E A D D F A D E F A E A B A E A C A E A D A E A E A E A F A E B C A E B D A E B E A E B F A E C D A E C E A E C F A E D E A E D F A E E F A F A B A F A C A F A D A F A E A F A F A F B C A F B D A F B E A F B F A F C D A F C E A F C F A F D E A F D F A F E F B C A B B C A C B C A D B C A E B C A F B C B C B C B D B C B E B C B F B C C D B C C E B C C F B C D E B C D F B C E F B D A B B D A C B D A D B D A E B D A F B D B C B D B D B D B E B D B F B D C D B D C E B D C F B D D E B D D F B D E F B E A B B E A C B E A D B E A E B E A F B E B C B E B D B E B E B E B F B E C D B E C E B E C F B E D E B E D F B E E F B F A B B F A C B F A D B F A E B F A F B F B C B F B D B F B E B F B F B F C D B F C E B F C F B F D E B F D F B F E F C D A B C D A C C D A D C D A E C D A F C D B C C D B D C D B E C D B F C D C D C D C E C D C F C D D E C D D F C D E F C E A B C E A C C E A D C E A E C E A F C E B C C E B D C E B E C E B F C E C D C E C E C E C F C E D E C E D F C E E F C F A B C F A C C F A D C F A E C F A F C F B C C F B D C F B E C F B F C F C D C F C E C F C F C F D E C F D F C F E F D E A B D E A C D E A D D E A E D E A F D E B C D E B D D E B E D E B F D E C D D E C E D E C F D E D E D E D F D E E F D F A B D F A C D F A D D F A E D F A F D F B C D F B D D F B E D F B F D F C D D F C E D F C F D F D E D F D F D F E F E F A B E F A C E F A D E F A E E F A F E F B C E F B D E F B E E F B F E F C D E F C E E F C F E F D E E F D F E F E F

In reply to Re: Generalizing Code: Generating Unique Permutations by Anonymous Monk
in thread Generalizing Code: Generating Unique Permutations by Anonymous Monk

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