Here's another iterator solution that avoids the recursion altogether. There are three differences:

• It will return undef instead of an empty array ref if the number of items is less than the amount to choose (ie 3 choose 5).
• The amount to choose is passed before the list of items instead of afterwards because I thought it was weird, although your way does make sense if one thinks of it like "Here are 50 things, choose 5".
• The output order is a bit different.

All three of these can certainly be fixed, but I wanted to keep the code clear.

sub choose_n_iter {
    my @todo = [ shift, [ @_ ], [] ];
    return sub {
        while ( @todo ) {
            my ( $n, $pool, $tally ) = @{ shift @todo };

            return $tally if $n == 0;

            next if @$pool == 0;
            
            my ( $first, @rest ) = @$pool;

            push @todo, [ $n - 1, \@rest, [ @$tally, $first ]],
                        [ $n    , \@rest,    $tally          ];
                        
        }
        return;
    };
}
[download]

And here is a more effecient version:

sub choose_n_iter {
    my @todo = [ shift, [ @_ ], [] ];
    return sub {
        while ( @todo ) {
            my ( $n, $pool, $tally ) = @{ shift @todo };

            return    $tally           if $n == 0;
            return [ @$tally, @$pool ] if $n == @$pool;

            next if @$pool == 0;
            
            my ( $first, @rest ) = @$pool;

            push @todo, [ $n - 1, \@rest, [ @$tally, $first ]];
            push @todo, [ $n    , \@rest,    $tally          ] if @res
+t;
                        
        }
        return;
    };
}
[download]

Thanks goes out to HOP for the recursion-to-iterator help.