Your intuition is correct.
Consider a system with a fixed number $N$ of nonrelativistic particles, all fermions. Ignoring spin for simplicity, the wavefunction of such a system is a function of $N$ points in space:
$$
\newcommand{\bfx}{\mathbf{x}}
\psi(\bfx_1,\bfx_2,...,\bfx_N).
\tag{1}
$$
If the $j$th and $k$th particles are the same species, then the wavefunction must satisfy
$$
\newcommand{\bfx}{\mathbf{x}}
\psi(\bfx_{\pi(1)},\bfx_{\pi(2)},...,\bfx_{\pi(N)})
=
-
\psi(\bfx_1,\bfx_2,...,\bfx_N)
\tag{2}
$$
for the permutation $\pi$ that exchanges $j\leftrightarrow k$ and leaves the other points unchanged. If the $j$th and $k$th particles are not the same species, then no such (anti)symmetry is required. In particular, if we have $5$ particles ($N=5$) all of different species, then the wavefunction doesn't need to have any (anti)symmetry at all. The fact that the particles are all fermions is irrelevant in that case. Nothing would change if they were bosons — in the strictly nonrelativistic and spinless model that we're using here for simplicity. (In a relativistic model, we can't ignore spin, and the number of particles is usually ill-defined, but I won't go into those complications here.)
We can consider all values of $N$ simultaneously using the formalism of creation/annihilation operators. Still ignoring spin for simplicity, we can describe a system of strictly nonrelativistic fermions using $K$ different creation operators $a_k^\dagger(\bfx)$ for each $\bfx$, with $k\in\{1,2,...,K\}$, where $K$ is the number of different species. If $|0\rangle$ is the state with no particles, then
$$
\int_{\bfx,\bfx'} \psi(\bfx,\bfx')
a_1^\dagger(\bfx)a_1^\dagger(\bfx')|0\rangle
\tag{3}
$$
is an example of a state with two particles of the same species, and
$$
\int_{\bfx,\bfx'} \psi(\bfx,\bfx')
a_1^\dagger(\bfx)a_2^\dagger(\bfx')|0\rangle
\tag{4}
$$
is an example of a state with two particles of different species. The assertion that the particles are all fermions can be expressed mathematically by the requirement that all of the creation operators anticommute with each other:
$$
a_j^\dagger(\bfx)a_k^\dagger(\bfx')
=
-a_k^\dagger(\bfx')a_j^\dagger(\bfx).
\tag{5}
$$
For particles of the same species ($j=k$), this immediately implies the Pauli exclusion principle: only the antisymmetric part of $\psi$ matters in (3), because the minus sign in (5) eliminates any contribution from the symmetric part. This follows from the fact that in (3), exchanging the subscripts is the same as exchanging the points. But for particles of different species ($j\neq k$), that's no longer true, and the minus sign in (5) has no consequence in thiscase. In fact, the minus sign in (5) can be eliminated when $j\neq k$ using a Klein transform.
Consider a system of $N = \sum_i n_i$ identical fermions, with the discrete hydrogen spectrum. In this case the energy levels are, in Coulomb units, well known to be $E_n = - 1/2n^2$ or, since I used "i" to refer to the $i = 1,2,...$'th energy level:
$$E_{i} = - \frac{1}{2i^2},$$
so that the total energy is $E = \sum_i n_i E_i$ where there are $n_i$ particles in the $i$'th energy level. The degeneracy $g_i$ of the $i$'th level (ignoring spin) is $i^2 = \sum_{l=0}^{i-1} (2l+1)$ degenerate states (i.e. different wave functions with different values of $(m,l,n)$). If we include spin it's now just $g_i = 2i^2$. In other words, in the $i$'th energy level $E_i$ there are $g_i = 2i^2$ different sttes/wave-functions $\psi_{(n_i,l_i,m_i,s_i)}(r,\theta,\phi)$ that the $n_i$ particles, with $n_i \leq g_i$, could be in.
Assume that the $n_i$ particles in the $i$'th energy level are an independent subsystem of the total system. This vitally important statement means that even though there are a bunch of different energy levels $E_1, E_2, ...$ we just focus on one energy level $E_i$. For all intent and purposes, this is our whole system for the moment, we deal with other $E_j$'s after.
The question is: in how many ways can we put the $n_i$ particles with energy $E_i$ into the $g_i$ degenerate states each with energy $E_i$ such that only one particle can fit into a given state with energy $E_i$, noting the total energy of the particles in this energy level is $n_i E_i$. This 'combinatorial' number is denoted $w(n_i,g_i)$. Once we calculate this for $E_i$, then, since the levels are all independent, the total statistical weight where we factor in all independent energy levels $E_1,E_2,...$ is just the product of all the separate $w(n_i,g_i)$'s: $W = \Pi_i w(n_i,g_i)$.
There are $g_i$ choices for the first particle, $g_i - 1$ for the second since it can't occupy the same state the first particle occcupied, etc... giving $g_i (g_i -1) ...(g_i - n_i + 1) = \frac{g_i!}{(g_i - n_i)!}$ but this over-counts as each permutation of the $n_i$ particles is treated as independent but they are identicaal particles so we divide by $n_i!$ to get
$$w(n_i,g_i) = \frac{g_i!}{n_i!(g_i-n_i)!}$$ for all $j$'s.
This is the statistical weight for the $E_i$ energy level. We can now repeat this for all $E_j$'s immediately finding $w(n_j,g_j) = \frac{g_j!}{n_j!(g_j-n_j)!}$.
For example, say $i = 4$ and that there are $n_i = n_4 = 3$ particles with this energy $E_4 = - \frac{1}{2 \cdot 4^2} = - \frac{1}{32}$ so that $n_i E_i = 3 E_4 = - \frac{3}{32}$ is the total energy of the $i=4$'th (sub)system. There are $g_4 = (\frac{1}{2}2+1) \sum_{l=0}^{n-1} \sum_{m=-l}^l 1 = 2 \cdot 3^2 = 18$ different wave functions $\psi_{(i=4,l_4,m_4,s_4)}(r,\theta,\phi)$ where $s_4 = \pm \frac{1}{2}$, $m_4$ goes from $-l_4$ to $l_4$, and $l_4$ goes from $0$ to $4-1 = 3$. Since they all correspond to the same energy eigenvalue they are called degenerate states, and again notice we set $n_4 = 3 \leq g_4 = 18$. Thus $w(n_i = 4,g_i = 18) = \frac{18!}{4!(14)!}$.
We can repeat this for every energy level $E_1,E_2,...$ in the Hydrogen atom spectrum even for ones with $n_k = 0$ i.e. ones with no particles in them, $w(n_k,g_k)=\frac{0!}{0!0!} = 1$ will hold for them. Since the levels are all independent, the total statistical weight is then just the product of all the $w(n_i,g_i)$'s: $W = \Pi_i w(n_i,g_i)$.
You'll notice we're supposed to know the values of $n_1,n_2,...$ also. Of course we're really going to try to treat them as variables and then maximize $W$ in terms of them, i.e. find the choice of $n_1,n_2,...$ which maximizes $W$, but it's up to us to choose them/have them handed to us, unless we're maximizing so we can then select this 'best' choice.
Hopefully you see your issue with spin is factored into and affect the value that $g_i$ takes, where it was $n_i^2$ or $2n_i^2$. This assumes spin $1/2$, if you changed the spin to $2J+1$ it would just change the value of $g_i$. The point in this calculation is just that if they are fermions you assume they can't occupy the same state depending on how the state is defined, whether it includes the spin or not.
Best Answer
“Two particles are said to be identical if all their intrinsic properties (mass, spin, charge, etc.) are exactly the same.” (Cohen-Tannoudji et al.) Two particles are indistinguishable if you can’t tell which is which, and that depends on the system. In classical mechanics, two identical particles may be distinguishable by their past histories, e.g. you can follow their paths without losing track of which is which. In quantum mechanics, identical particles are indistinguishable, but identical particles may be in distinguishable states, and it is easy to confuse particles and states. Two neutral 6Li atoms each have identical constituents (3 electrons, 3 neutrons, 3 protons) but these constituents can be arranged in different internal states that do not all have the same spin or mass. Two 6Li atoms are only identical if they are in the same internal state.
Even if particles have identical intrinsic properties, they can be in different external states as part of larger system. In a magnetic field, identical 6Li atoms will be in one of two distinguishable states with different energies, depending on whether the atom spin is aligned or anti-aligned with the magnetic field. I believe that in the paper you reference, the “indistinguishable” pairs of atoms are those with aligned spins, and “distinguishable “ pairs are those with opposite spins. This is compatible with how we might think about the two electrons in a neutral helium atom. If they have different spins then they can both be in the 1S level since the spin-up state is distinguishable from the spin-down state. If they have the same spin then they are indistinguishable in the 1S level, so by the Pauli exclusion principle one electron must be in a higher energy level.
Spin-projection to an outside axis is not an intrinsic property of a particle, it is a parameter describing its external quantum state. Internal parameters such as the rest mass, charge, and spin of all electrons are always the same for identical particles, but external state parameters can change. Just because two electrons can have different positions, momenta, or spin-projections does not mean that all electrons are not identical, it just means they can be in different external states. $F$ is an internal quantum number, and the 6Li $F=\frac{1}{2}$ and $\frac{3}{2}$ states have different masses and total angular momenta and hence are different particles. Unlike the $m_s$ and $m_l$ quantum numbers that parameterize differences in internal atomic structure due to the possible relative orientations of nuclear and electron orbits and spins, $m_F$ can only be defined relative to an external axis, and the rest mass, charge, and spin of the atom are independent of $m_F$. So two $2 S_{1/2}$ $F=\frac{1}{2}$ 6Li atoms are identical particles, but they may be in different $m_F$ states.
The answers to What are the differences between indistinguishable and identical? and Distinguishing identical particles, may also be helpful.