I started with GrandFather's code, rewriting it slightly, then dumped the counts of a series with constant $holes, then looked it up on the OEIS. Hubris, I know.

Looks like the solution is a simple lookup in binomial coefficients (on a diagonal in the Pascal's triangle).

#! /usr/bin/perl -l

my $cards        = 12;
my $holes        = 7;

sub arrange {
    my ($n, $k, $i, $prefix) = @_;
    return "$prefix $n" if ++$i == $k;
    map arrange($n - $_, $k, $i, "$prefix $_"), (1 .. $n - ($k - $i));
}

sub solve {
    my ($n, $k) = (shift, shift);
    return [@_, $n] unless (@_ - $k + 1);
    map solve($n - $_, $k, @_, $_), 1 .. $n + (@_ - $k + 1);
}

# binomial coefficients ie combinations C(n,k)
# (this can be written far more efficiently, of course)
sub choose {
    my ($n, $k) = (shift, shift);
    return 0 if $k < 0 || $k > $n;
    !$n || choose($n-1, $k) + choose($n-1, $k-1)
}

print "@$_" for solve($cards, $holes);
print "== ", int(()=solve($cards, $holes));

print "@{[map {int (()=solve($holes + $_, $holes))} 0..17]}";
print "@{[map {choose($holes -1 + $_, $_)} 0..17]}";
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Update. Here's a solution with Algorithm::Combinatorics.

#! /usr/bin/perl -l

use Algorithm::Combinatorics ':all';

my $cards = 12;
my $holes = 7;

my $iter = combinations_with_repetition( [1 .. $holes], $cards - $hole
+s );

while (my $x = $iter->next) {
    print "@{distrib($holes, $x)}";
}

sub distrib {
    my @d = (1) x shift;
    ++$d[$_-1] for @{+shift};
    return \@d;
}
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