There are 12 permutations of this object that transform it onto itself that are unit and in the form of $(1\ 2\ 3\ 4\ 5\ 6)$; $(7\ 8\ 9\ 10\ 11\ 12)$; $(1\ 3\ 5)(2\ 4\ 6)(7\ 9\ 11)(8\ 10\ 12)$; $(1\ 4)(3\ 6)(2\ 5)(3\ 6)(7\ 10)(8\ 10)(9\ 12)$; $(1\ 5\ 3)(2\ 6\ 4)(7\ 11\ 9)(8\ 12\ 10)$; $(1\ 6\ 5\ 4\ 3\ 2)(7\ 12\ 11\ 10\ 9\ 8)$; and $(1\ 8)(2\ 7)(3\ 12)(4\ 11)(5\ 10)(6\ 9)$ etc. Exactly one of them has 12, 2 of them have 2, 2 of them have 4 disjoint cycles, and lastly 7 of them are a product of 6 disjoint cycles. Therefore, according to the orbit counting lemma, $t = \frac{1}{12}(1 \cdot 2^{12} + 2 \cdot 2^2 + 2 \cdot 2^4 + 7 \cdot 2^6)$ where $t$ is the number of orbits. In other words, the total number of pavements that are damaged in a different manner equals $4584/12 = 382$. Thus, since $2009 < 382 \cdot 6$, it is not possible to select the required 7 pavements with the same damage out of 2009 pavements.