The paramagnetism of materials can be explained by the behavior of the alignment or not of the magnetic dipole moments of their atoms. Simply put, in a convention where the Cartesian axis $z$ points up, the magnetic dipole moment $z $ component can assume only the values + $\mu_0$ (up) or $-\mu_0 $ (down), where $\mu_0$ is the intensity of the magnetic dipole moment of each atom. In the absence of external magnetic field, the magnetic dipole moments of the atoms are randomly oriented, so that the magnetization $M$ of the system, given by the sum of the magnetic dipole moments, is zero. In the presence of a magnetic field pointing in the $z$ direction, $M$ is nonzero, since magnetic dipoles tend to align with the external field. However, the alignment is not total, because besides the interaction with the field, there are thermal interactions that favor the disorder.
In paramagnetic materials, the intensity of the external field $B$ and the temperature $T$ determine the probability of alignment of a magnetic dipole with the field. Consider a particular situation in which the probability that a magnetic dipole will orientate in the field direction is $p = 3/4$. What is the probability that a group of 6 atoms will have magnetization $M = 2\mu_0$?
The way I thought to ask this question was: First there are two possibilities for atoms: being oriented in the field direction ($3/4$) or not being ($1/4$).
Then there are two more possibilities: Pointed up ($1/2$) or down ($1/2$). There are then 3 ways to organize them:
$$\begin{align} ++++ -- & \Rightarrow 2+ \\ +++ - 00 & \Rightarrow 2+ \\ ++ 0000 & \Rightarrow 2+ \\ \end{align}$$
Then the probability would be:
$$\left(\frac{3}{4}\right)^6 \left(\frac{1}{2}\right)^6 \binom 62 +\left(\frac{3}{4}\right)^4\left(\frac{1}{4}\right)^2 \left(\frac{1}{2}\right)^4 \binom 62 +\left(\frac{3}{4}\right)^2\left(\frac{1}{4}\right)^4 \left(\frac{1}{2}\right)^2\binom 62 $$
I think you've misunderstood the physics aspect of this. The field direction isn't a further direction in addition to up and down; the field is oriented in the $z$ direction. The question is asking you to find the probability of $4$ of the $6$ dipoles being oriented up (in the field direction) and $2$ down (opposite). This is
$$ \binom64\left(\frac34\right)^4\left(\frac14\right)^2=\frac{1215}{4096}\approx0.3\;. $$