Why is the change in entropy is different for different interpretations?
From the perspective of Statistical Mechanics:

Multiplicities do depend on the external parameters that specify the energy level of the system.

For ideal gas of $N$ indistinguishable molecules, the multiplicity is given by $$\Omega= \frac{1}{N!}\frac{V^N}{h^{3N}}\,\frac{2\pi^{3N/2}}{(3N/2- 1)!}(\sqrt{2mE})^{3N-1}$$

So, indeed the multiplicity increases when the position space increases.

When the external parameter like volume of the container is increased, the energy levels come closer and then the particles can have more microstates for the same $E$.

Prior to the volume increase, the microstates of the system are $\Omega_{N,E,V_i}\;_;$ when the volume is increased, not will be there the microstates $\Omega_{N,E,V_i}$ still be available after the removal of the constraint, but also, there be addition of microstates. In general, $$\Omega_{V_f}\geq \Omega_{V_i}\;.$$

So, as the volume increases, the multiplicity increases and so does the entropy.

As requested by OP:

From the perspective of Thermodynamics:

Suppose the two dissimilar indistinguishable gases $A$ and $B$ having the same number of particles, energy, volume are partitioned from each other.

Now, when the partition is removed, the energy of the gas $A$ confined to a localised area gets dispersed over a greater volume. 

Thus the entropy of gas $A$ increases and that is given by \begin{align}\Delta S_A &amp;= Nk\,\ln \frac{V_f}{V_i}\\ &amp;= Nk\,\ln \frac{2V}{V}\\ &amp;= Nk\,\ln 2\;.\end{align}

Same happens for gas $B$ also.

Therefore, the change in entropy of the system is given by \begin{align}\Delta S_\textrm{system}&amp;= \Delta S_A + \Delta S_B\\ &amp;= 2Nk\ln 2 \gt 0\end{align}

Therefore, entropy increases when the partition is removed.