Fundamental particles, like electrons are all identical or indistinguishable.
We define $\psi(\underline{r}_1,\underline{r}_2)$ to be the wavefunction of a two-electron system where one electron is at $\underline{r}_1$ and the other at $\underline{r}_2$. We can write that
$$ \Bigl|\psi(\underline{r}_1,\underline{r}_2)\Bigr|^2=\Bigl|\psi(\underline{r}_2,\underline{r}_1)\Bigr|^2 $$
if they equivalent the probability of the first to be at $\underline{r}_1$ and the second at $\underline{r}_2$ must be equal to the probability for the first to be at $\underline{r}_2$ and the second at $\underline{r}_1$.
Thus we can write
$$ \psi(\underline{r}_1,\underline{r}_2)=\mathrm{e}^{i\alpha}\psi(\underline{r}_2,\underline{r}_1) $$
We define an operator that corresponds to swapping the properties of the two particles (position + spin)
$$ \widehat{P}\,\psi\bigl(\{\underline{r}_1,s_1\},\{\underline{r}_2,s_2\}\bigr)=\psi\bigl(\{\underline{r}_2,s_2\},\{\underline{r}_1,s_1\}\bigr) $$
where $\widehat{P}$ has eigenvalues $\pm 1$
$$ \psi\bigl(\{\underline{r}_1,s_1\},\{\underline{r}_2,s_2\}\bigr)=\pm\psi\bigl(\{\underline{r}_2,s_2\},\{\underline{r}_1,s_1\}\bigr) $$
For which we use the short hand notation
$$ \psi(1,2)=\pm\psi(2,1)\, $$
🗒️ Note: the operator $\widehat P$ is sometimes called the spin-parity operator or the particle exchange operator
A given type of particle always has the same sign under particle exchange
Particles that have
$$ \psi(1,2)=+\psi(2,1) $$
are called bosons ( photons or helium-4 atoms) they have integer spin $s$ ($s=1$ for photon)
Particles that have
$$ \psi(1,2)=-\psi(2,1) $$
are called
fermions (electrons or helium-3 atoms) they have half-integer spin ($s=\frac 12$ for electron)
Ignoring spin suppose some system has single-particle wavefunctions
$$ \psi_\alpha(\underline{r}) \quad\text{and}\quad \psi_\beta(\underline{r}) $$
where $\alpha \ne \beta$ are sets of quantum numbers fully describing the state
Separation of variables doesn’t work so we try the combinations
$$ \begin{aligned} \psi^{(S)}(\underline{r}_1,\underline{r}2) &=& \frac1{\sqrt{2}}\Bigl( \psi\alpha(\underline{r}1)\,\psi\beta(\underline{r}2)+ \psi\alpha(\underline{r}2)\,\psi\beta(\underline{r}_1)\Bigr) \\ \psi^{(A)}(\underline{r}_1,\underline{r}2) &=& \frac1{\sqrt{2}}\Bigl( \psi\alpha(\underline{r}1)\,\psi\beta(\underline{r}2)- \psi\alpha(\underline{r}2)\,\psi\beta(\underline{r}_1)\Bigr)\end{aligned} $$
Where $S$ and $A$ stand for symmetric and antisymmetric respectively, and the factor of $1/\sqrt{2}$ means that is $\psi_{\alpha,\beta}$ are normalized $\psi^{(S,A)}$ are too
if $\alpha=\beta$ then $\psi^{(S)}(\underline{r}_1,\underline{r}2)=\psi\alpha(\underline{r}1)\,\psi\alpha(\underline{r}_2)$ is allowed but $\psi^{(A)}(\underline{r}_1,\underline{r}_2)=0$ there is no antisymmetric combination possible
Identical bosons are allowed to be in the same state. In fact they like to be e.g Bose-Einstein condensation, which leads to superfluidity of liquid helium, or stimulated emission of photons in a laser
<aside> 🐞 Pauli Exclusion principle: Identical fermions are not allowed to be in the same state. Each electron state is either occupied or unoccupied but never multiply occupied
</aside>