Sure. No need to do any recursion here. Just count in binary. Here is a version that creates a closure (an anonymous subroutine that holds the needed data inside of itself -- sort of like a tiny "object") that returns the next combination each time it is called:

sub combinations {
    my @list= @_;
    my @pick= (0) x @list;
    return sub {
        my $i= 0;
        while( 1 < ++$pick[$i]  ) {
            $pick[$i]= 0;
            return   if  $#pick < ++$i;
        }
        return @list[ grep $pick[$_], 0..$#pick ];
    };
}
my $next= combinations( 50..59 );
my @comb;
while(  @comb= $next->()  ) {
    # do work with @comb here
}
# Note that the empty set is a valid combination but is
# the last combination returned which also indicates "no
# more combinations left.  So the above loop doesn't bother
# processing the empty list.  If you want to process the
# empty set, then use:
my @comb;
do {
   # do work with @comb here
} while(  @comb= $next->() );
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Update: My code finds combinations but the original code finds permutations even though the author asked for combinations. (See (tye)Re: Permutations if you don't know the difference between the two.)

Of course, my favorite way of finding permutations is Permuting with duplicates and no memory.