Dear Josh, the wave functions are only perfectly symmetric and/or perfectly antisymmetric - each of the factors is - in the case of two particles. As Fabian correctly says, for more than 2 particles, the wave function isn't perfectly symmetric and isn't perfectly antisymmetric with respect to particular transpositions of the two particles. The general character of the wave function's behavior is given by a Young diagram.
That's why what you want to be proved can only be proved for two particles. In the text below, I will therefore assume that there are just two electrons. What you want to be proved is that if the total $L=L_1+L_2$ of the two electrons is even, the wave function is even under the exchange of the two electrons, and it's odd if $L$ is odd.
It's not hard to see. By rotational symmetry, the states with a given value of $L$ form a multiplet with $2L+1$ components because $L_z$ may go from $-L$ to $+L$ with the spacing one. Without the loss of generality, you may focus on the wave function with the maximum value of $L_3$, namely $L_z=L$.
To get this maximum value, you need $L_z=L_{z,1}+L_{z,2}$ to be composed of two equally maximal, equally oriented terms. Because $L_{z,1}$ and $L_{z,2}$ also go between $-L_1$ and $+L_1$, or similarly between $-L_2$ and $+L_2$, it's clear that the only way to get $L_z=L$ is to have $L_{z,1}=L_1$ and $L_{z,2}=L_2$: both $z$-components must be maximum, too. But if it is so, then the wave function is simply
$$ \psi_{L_3=L} = \psi_{L_{z,1}=L_1} \otimes \psi_{L_{z,2}=L_2} $$
For $L_1=L_2$, you can easily see that this wave function has to be symmetric under the permutation of $1$ and $2$. After all, it is the tensor product of the two equal pieces. If you antisymmetrized it, you would get zero. And in fact, the total wave function cannot have any complete symmetry or complete antisymmetry under the exchange of the particles $1,2$ if $L_1\neq L_2$. It's because $L_1^2$ acting on the total wave function gives you $L_1(L_1+1)$ times the wave function, while $L_2^2$ acting on the total wave function gives you $L_2(L_2+1)$. Because the two eigenvalues are not equal for $L_1\neq L_2$, the total wave function can't be symmetric under the exchange of $1,2$.
Again, the question about the symmetry of the orbital wave function only has a sharp answer is there are two particles and if they have the same $L_1=L_2$ - and in that case, the function is symmetric under the permutation.
Similarly, one can prove that the exchange of the two particles with the same half-integer spins $S_1=S_2$ produces a minus sign - assuming that $S_1=S_2$ differs from an integer by $1/2$. To do such things, it's useful to imagine that the components of the multiplet with a given $J$ are organized into a symmetric spintensor.
All the $2J+1$ components of the multiplet with the angular momentum $J$ may be expressed as a completely symmetric "tensor" with 2-valued spinor indices,
$$ T_{abc\dots z} $$
Each index is either $0$ or $1$. The number of ones goes from $0$ to the number of indices $N$ - so there are $N+1$ components of this tensor.
Because each index brings $1/2$ to the total angular momentum, it's clear that the number of indices is $N=2J$. Indeed, then we have $N+1=2J+1$ components.
If there were pairs of indices that are antisymmetric with respect to the exchange of the two indices, one could factorize $\epsilon_{ab}$, a totally antisymmetric object. So only the total symmetry is relevant for our simplest case. Now, the spintensor for an angular momentum $J=J_1+J_2$ object may be written as the symmetrization of
$$ T_{abc\dots z} = T^{1}_{(abc\dots m} T^{2}_{nop\dots z)} $$
where the parentheses represent the complete symmetrization - because the multiplet is represented by the totally symmetric tensor, as we said. The tensors $T^1$ and $T^2$ correspond to the $J_1$ and $J_2$ pieces.
However, if the total angular momentum $J$ differs by an odd number from its maximum value $J_1+J_2$, like in the case of the singlet where $J_1=J_2=1/2$ but $J=0$, then we must factorize those epsilons.
$$ T_{abc\dots z, \,\,{\rm missing\,\,}{mn}} = T^{1}_{(abc\dots l} T^{2}_{op\dots z)} \epsilon^{mn} $$
The epsilon was added to the right hand side to reduce the number of indices by two - i.e. the total spin by one. Because the epsilon is antisymmetric, it changes the symmetry of the whole $T$ under the exchange of the two groups of indices. Each time you reduce the total $J$ by one, the symmetry property changes from symmetry to antisymmetry or vice versa. So the sign obtained from the permutation is given, for $J_1=J_2$, by $(-1)^{J-J_1-J_2}$.
It's also useful to know that the orbital wave functions of a single particle that can be expressed as spherical harmonics $Y_{lm}$ pick the factor of $(-1)^{l}$ under parity i.e. the factor of $(-1)^{l+m}$ under $\theta\to\pi-\theta$.
The valence quark content of a baryonic state is $qqq$, that is to say three identical quarks make up the content of the bound state amongst a sea of partons. Each quark therein carries a series of quantum numbers which serve to denote the different possible states of the quark and, collectively, must be such that a simultaneous permutations of the degrees of freedom yields an overall antisymmetric state, in accordance with the fact the quantum state must obey Fermi-Dirac statistics.
For non excited states (the so called lowest lying states), the spatial component of the direct product is always symmetric ( $S$-wave orbital angular momentum) and colour is antisymmetric so the combination $|\text{spin} \rangle \otimes |\text{flavour} \rangle$ must be symmetric. The observable states transform under irreducible multiplets of (approximate) $SU(3)$ flavour symmetry and so the '$|uds \rangle$' state is actually a flavour symmetric combination in the $10$ decuplet, a flavour antisymmetric combination in the $1$, and appears with mixed flavour symmetries in the remaining two octets in agreement with the group theoretic decomposition $3 \otimes 3 \otimes 3 = 1 \oplus 8 \oplus 8 \oplus 10$.
So. e.g for the totally flavour symmetric combination, the spin state of the wavefunction must be in one of the symmetric spin $3/2$ states.
Best Answer
The wavefunction is always antisymmetric. It does not matter if they have similar, different, or even identical numbers for $n$ and $l$.
But let's be clear. The wavefunction is a function from the configuration space into the joint spin state. So for a spin 1/2 particle the wavefunction is a function from $\mathbb R^3$ to $\mathbb C^2$ whereas for two spin 1/2 particles the wavefunction is a function from $\mathbb R^6$ to $\mathbb C^2\otimes\mathbb C^2$ and it has to be antisymmetric if they are indistinguishable fermions, symmetric if they are indistinguishable bosons, and no restrictions (could be symmetric, antisymmetric, or neither) if they are distinguishable.
And by the symmetries we mean things like $\Psi(\vec a,\vec b)=-\Psi(\vec b,\vec a).$
So let's do an example. If $\psi_{nlm}:\vec r\mapsto\psi_{nlm}(\vec r)$ is a solution with a particular value of $n,$ $l,$ and $m$ for the scalar (spin zero) Schrödinger equation for a 1/r potential then you could make a spin 1/2 solution by having $$\Psi:\mathbb R^3\rightarrow\mathbb C^2:\vec r\mapsto\psi_{nlm}(\vec r)\left[\begin{matrix}\alpha\\ \beta\end{matrix}\right].$$
For a pair of indistinguishable spin 1/2 particles (e.g. two electrons, two muons, etcetera) a possible wavefunction is
$$\Psi:\mathbb R^6\rightarrow\mathbb C^2\otimes\mathbb C^2,$$ $$\Psi:(\vec a,\vec b)\mapsto\psi_{nlm}(\vec a)\psi_{NLM}(\vec b)\left[\begin{matrix}\alpha\\ \beta\end{matrix}\right] \otimes\left[\begin{matrix}A\\ B\end{matrix}\right]-\psi_{nlm}(\vec b)\psi_{NLM}(\vec a)\left[\begin{matrix}\alpha\\ \beta\end{matrix}\right] \otimes\left[\begin{matrix}A\\ B\end{matrix}\right].$$
In fact you can build more complicated states out of these antisymmetric states. But the note complicated states will remain antisymmetric.