Carbon has a $\mathrm{p^2}$ configuration, and within the Russell-Saunders coupling scheme, we have
$$\begin{align} s_1 = s_2 &= \frac{1}{2} & S &= 1,0 \\ l_1 = l_2 &= 1 & L &= 2,1,0 \\ \end{align}$$
which gives rise to $^3D$, $^1D$, $^3P$, $^1P$, $^3S$, and $^1S$ terms. Out of these six, only $^1D$, $^3P$, and $^1S$ are allowed by the Pauli principle.
It's not hard to show that the $^3D$ term is forbidden, since this term has to contain one state with $m_{l1} = m_{l2} = 1$, which would correspond to a spatial wavefunction that is symmetric with respect to interchange of the two electrons. Pairing this with a triplet spin wavefunction, which is also symmetric, would lead to an overall symmetric wavefunction, which is not allowed.
But that's all the textbooks say - they don't describe why the $^1P$ term is forbidden, for example. I know how to construct a table of microstates, and I know I can use the direct product table of the full rotation group to find that $P \times P = S + [P] + D$, and I know that both of these methods show that the allowed term symbols are only $^1D$, $^3P$, and $^1S$.
Essentially, I am looking for a more intuitive explanation of why the $^1P$ and $^3S$ states are forbidden that preferably links back to the requirement that the wavefunction be antisymmetric - if that is possible.
Alternatively, an explanation of why $P \times P = S + [P] + D$ would also work. I know that from the Clebsch-Gordan series we get $S$, $P$, and $D$ terms, but not why the $P$ state should be antisymmetric. I do know the general idea of how to obtain the vector coupling coefficients (by repeated application of raising/lowering operators), and these do show that the $P$ state is antisymmetric, but I'd like a more interpretative answer than just "that's how the maths works out".