Finding the symmetry of electron spin and position states using Clebsch-Gordan tables? I know that spin and position states must be symmetric and antisymmetric (and vice versa), but I can't figure out how to use Clebsch-Gordan Tables to figure out the symmetry of either one.
For example, Carbon's ground state has a total spin of 1 due to Hund's first rule. How do I know that this spin state is symmetric?
Also, how do I figure out if the position states L = 1 and L = 2 are symmetric, so I can figure out which is correct for the ground state of Carbon?
 A: Carbon has two valence electrons in the p-shell.  To obtain total spin $S=1$ the spin wavefunction must be symmetric.  The best way to see this is that the $S=1$ states will contain a $M_s=1$, which must be unique product $\vert 1/2,+\rangle_1\vert 1/2,+\rangle_2$, clearly symmetric under permutation. 
As a result, the spatial part of the wavefunction must be antisymmetric.  In the decomposition of $(\ell =1)\otimes(\ell =1)=(L=2)\oplus (L=1)\oplus(L=0)$, it is the $L=1$ part that is antisymmetric.  One can reach this argument by counting knowing that there must be $6$ symmetric states and $3$ antisymmetric states.  
The counting argument starts by noting that the $L=2,M_L=2$ state must be the unique product state $\vert 2,2\rangle=\vert 1,1\rangle_1\vert 1,1\rangle_2$.  Thus, all $5$ $L=2$ states are fully symmetric.  Since the total number of symmetric states is $6$, the one remaining state is the $L=0$ state.  This leaves the three $L=1$ states as antisymmetric.
In terms of Clebsch-Gordan coefficients, antisymmetry is visible because the coefficients $C_{\ell_1m_1;\ell_2m_2}^{LM}$ for $\ell_1=\ell_2=L=1$, pick up an overall sign $(-1)^{\ell_1+\ell_2-L}=-1$ when $(\ell_1m_1)$ and $(\ell_2m_2)$ are interchanged, i.e. 
$$
C_{\ell_1m_1;\ell_2m_2}^{LM}=-C_{\ell_2m_2;\ell_1m_1}^{LM}\, .
$$
This phase $(-1)^{\ell_1+\ell_2-L}$ is $+1$ for $L=2$ or $L=0$, confirming that these are symmetric states.

Edit: actually your initial statement is limited to two particles, for more generally it is not quite correct.  The strictly correct statement is that the spin and spatial part of the wavefunctions need not be symmetric or antisymmetric by themselves, but their product must be antisymmetric (for fermions at least).  
In systems with $3$ or more particles, it is possible for the spin or spatial wavefunctions to have more complicated symmetries than just full symmetry or antisymmetry.  The mathematics behind this are the representation of the permutation group of $n$ objects, with $n=3$ for the example of three particles.  It is known how to combined spin and spatial wavefunctions of mixed symmetry to obtain a fully antisymmetric total wavefunction.
