The construction of antisymmetric wave functions, using the Pauli antisymmetry principle, is very complex for atoms with more than two electrons. In these cases we use the Slater determinants, which generate antisymmetric functions and do not distinguish between electrons. To construct a Slater determinant, the spin-orbitals occupied by the electrons are placed in the columns, while each row corresponds to one electron.
As an example, we will start by constructing the ground state of lithium by using Slater determinants.
Step 1. Write the electron configuration, Li:$1s^22s^1$.
Step 2. Write the spin-orbitals, $1s\alpha$; $1s\beta$; $2s\alpha$; $2s\beta$
Step 3. Write the determinant by placing the spin-orbitals in the columns. The first row is for the electron (1), the second for the electron (2) and the third for the (3).

$$\frac{1}{\sqrt{6}}\left| \begin{array}{ccc} 1s(1)\alpha (1) & 1s(1)\beta (1) & 2s(1)\alpha (1)\\ 1s(2)\alpha (2) & 1s (2)\beta (2) & 2s(2)\alpha (2)\\ 1s(3)\alpha (3) & 1s(3)\beta (3) & 2s(3)\alpha (3) \end{array} \right|$$

In this determinant we do not use the spin-orbital $2s\beta$ since lithium only has three electrons. However, the electrons are found with equal probability in the orbital spins $2s\alpha$ and $2s\beta$, giving rise to a doubly degenerate ground state. The wave function of this second state is given by:

$$\frac{1}{\sqrt{6}}\left| \begin{array}{ccc} 1s(1)\alpha (1) & 1s(1)\beta (1) & 2s(1)\beta (1)\\ 1s(2)\alpha (2) & 1s (2)\beta (2) & 2s(2)\beta (2)\\ 1s(3)\alpha (3) & 1s(3)\beta (3) & 2s(3)\beta (3) \end{array} \right|$$

As a second example, we will construct the ground state of the helium atom, using Slater determinants, He:$1s^2$

$$\frac{1}{\sqrt{2}} \left| \begin{array}{cc} 1s(1)\alpha(1) & 1s(1)\beta(1)\\ 1s(2)\alpha(2) & 1s(2)\beta(2) \end {array} \right|$$

Expanding the determinant we can verify that it coincides with the one obtained using the Pauli antisymmetry.