The Correct Statement of the Third Law of Thermodynamics The Third Law of Thermodynamics can be stated in various ways, one of which is:

The entropy of a perfect crystal at absolute zero is exactly equal to
  zero.

Is this true for only "perfect crystals" and not for (say) fluids?
 A: The hamiltonian of a perfect crystal can be approximated at low temperature as the sum of harmonic oscillator hamiltonians. In 1D we have
$$H = \sum_{i=1}^N \frac{p_i^2}{2 m} + \frac 1 2 m \omega^2 \sum_{ij} ( r_i- r_j)^2 $$
where the $ij$ sum is over nearest neighbors. 
It is possible to verify that the eigenvalues of this hamiltonian are
$$ E_n = \left( \frac 1 2 + n \right) 2 \hbar \omega \left| \sin \left(\frac {ka}{2} \right)\right| $$
Where $n=0,1,2,3,\dots$, $k$ is the wavevector and $a$ is the lattice spacing. There is only one fundamental state, namely the one with $n=0$. 
If the fundamental state is only one, entropy must vanish. This is clear from the Boltzmann relation:
$$S= k_B \log (\Omega)$$
where $\Omega$ is the number of micostates. If there is only one possible microstate (the fundamental state), $S$ must be $0$ (because $\log(1)=0$).
But there are systems are conceivable which have more than one fundamental state, i.e. in which the fundamental state is degenerate. If the degeneracy is smaller than exponential, however, there is no real problem, since the entropy per particle $S/N$ still vanishes in the thermodynamic limit. For example, if the degeneracy is of order $N^n$ we have
$$\lim_{N\to \infty} \frac{S}{N} \propto \lim_{N\to \infty} \frac{\log(N^n)}{N} =  n \lim_{N\to \infty} \frac{\log(N)}{N} = 0 $$
The real problem is when the degeneracy of the ground state is exponential, since in this case we have
$$\lim_{N\to \infty} \frac{S}{N} \propto \lim_{N\to \infty} \frac{\log(a^N)}{N} = \log(a) \neq 0$$
So the "third law" of Thermodynamics fails in systems with an exponential number of ground states. 
When we (improperly) apply the definition of $S$ to non-equilibrium systems, like glasses, this phenomenon is known as residual entropy. 
This is why it is necessary to specify "of a perfect crystal" when stating the third law.
Then, as mentioned as John Rennie in the comments, there are also some exceptions, like liquid Helium, which does not crystallize as $T \to 0$, but it forms a Bose-Einstein condensate. Also in this case, there is only one ground state and therefore $S(T \to 0)=0$.
