In atomic spectroscopy, the electronic state of an atom or ion can be specified by its atomic term symbol

$$^{2S+1}L_J,$$

where $S$ is the total spin angular momentum, $L$ is the total orbital angular momentum and $J$ is the total angular momentum.

One method of determining all term symbols for an electron configuration is by constructing a table counting all of its possible microstates. For a given orbital quantum number $l$, there is a maximum of $n=2(2l+1)$ orbital positions for the electrons to occupy. The total number of possible microstates $N$ can be calculated as a combination of all electrons occupying the available orbital positions,

$$N = \left(\begin{array}{c} n \\ e \end{array}\right) = \frac{n!}{e!(n-e)!},$$

where $n$ is the number of orbital positions for a given $l$ and $e$ is the number of electrons.

For example, consider the electron configuration $2s 2p^2$. The total number of possible microstates can then be calculated as

$$N = N_{2s}\times N_{2p} = \frac{n_{2s}!}{e_{2s}!(n_{2s}-e_{2s})!} \times \frac{n_{2p}!}{e_{2p}!(n_{2p}-e_{2p})!} = 30,$$

where we made the substitutions $n_{2s}=2$, $e_{2s}=1$ to account for a single electron occupying the $2s$ orbital, and $n_{2p}=6$, $e_{2p}=2$ to account for two electrons occupying the $2p$ orbital. Therefore, we have 30 total microstates for the $2s 2p^2$ configuration.

From the list of all possible microstates, all possible combinations of $(M_L, M_S)$ can be counted and placed in a table. The multiplet term symbol is obtained by accounting for $J$ for each state. For the example of $2s 2p^2$, we have:

$$M_L \begin{array}{c|cccc} & &M_S & & \\ & -\frac{3}{2} & -\frac{1}{2} & \frac{1}{2} & \frac{3}{2} \\ \hline -2 & 0 & 1 & 1 & 0 \\ -1 & 1 & 3 & 3 & 1 \\ 0 & 1 & 4 & 4 & 1 \\ 1 & 1 & 3 & 3 & 1 \\ 2 & 0 & 1 & 1 & 0 \end{array}$$

We can then start extracting terms from this table, starting from the top-left corner to the bottom-right. From the top row, we can extract the term $^2D$ for $|M_L| = 2$ and multiplicity 2: one with $M_S = -1/2$ and one with $M_S=1/2$. Accounting for $J$, we obtain $^2D_{3/2,\,5/2}$. When subtracting, we remove 1 from the entire column. The resulting table is then:

$$M_L \begin{array}{c|cccc} & &M_S & & \\ & -\frac{3}{2} & -\frac{1}{2} & \frac{1}{2} & \frac{3}{2} \\ \hline -1 & 1 & 2 & 2 & 1 \\ 0 & 1 & 3 & 3 & 1 \\ 1 & 1 & 2 & 2 & 1 \end{array}$$

where we have removed the $M_L = -2$ and $M_L = 2$ rows since they now contain all 0s. In the next step, we can extract the term $^4P$ for $|M_L| = 1$ and multiplicity 4: one each with $M_S = -3/2, -1/2 , 1/2, 3/2$. Accounting for $J$, we obtain $^4P_{1/2,\,3/2,\,5/2}$. The resulting table is:

$$M_L \begin{array}{c|cccc} & &M_S & & \\ & -\frac{3}{2} & -\frac{1}{2} & \frac{1}{2} & \frac{3}{2} \\ \hline -1 & 0 & 1 & 1 & 0 \\ 0 & 0 & 2 & 2 & 0 \\ 1 & 0 & 1 & 1 & 0 \end{array}$$

Here we can extract the term $^2P$ for $|M_L| = 1$ and multiplicity 2: one each with $M_S = -1/2, 1/2$. Accounting for $J$, we obtain $^2P_{1/2,\,3/2}$. The resulting table is:

$$M_L \begin{array}{c|cccc} & &M_S & & \\ & -\frac{3}{2} & -\frac{1}{2} & \frac{1}{2} & \frac{3}{2} \\ \hline 0 & 0 & 1 & 1 & 0 \\ \end{array}$$

where we have removed the $M_L = -1$ and $M_L = 1$ rows since they now contain all 0s. We are left with the final row, which we can extract the term $^2S$ for $|M_L| = 0$ and multiplicity 2: one each with $M_S = -1/2, 1/2$. Accounting for $J$, we obtain $^2S_{1/2}$.

Therefore for the $2s 2p^2$ configuration, we have obtained the following 8 terms:

$${}^2D_{3/2,\,5/2}, {}^4P_{1/2,\,3/2,\,5/2}, {}^2P_{1/2,\,3/2}, {}^2S_{1/2}$$

I have written a python script to calculate atomic term symbols based on this method here.