The correct argumentation realises, that the particles are indistinguishable!
Before removing the separator:
$$
Z_{\mathrm{a}}=\frac{Z_{1}^{N}}{N !} \cdot \frac{Z_{1}^{N}}{N !}
$$
After removing the separator:
$$
Z_{\mathrm{b}}=\frac{\left(2 Z_{1}\right)^{2 N}}{(2 N) !}
$$
Before removing the separator: $F_{\mathrm{a}}=-k_{\mathrm{B}} T \ln \left(\frac{Z_{1}^{N}}{N !} \cdot \frac{Z_{1}^{N}}{N !}\right)=-k_{\mathrm{B}} T 2\left[N \ln Z_{1}-N \ln N+N\right]$ After removing the separator: $F_{\mathrm{b}}=-k_{\mathrm{B}} T \ln \left(\frac{\left(2 Z_{1}\right)^{2 N}}{(2 N) !}\right)=-k_{\mathrm{B}} T\left[2 N \ln 2 Z_{1}-2 N \ln (2 N)+2 N\right]$ and $\Delta F=-k_{\mathrm{B}} T 2 N\left[\ln 2+\ln Z_{1}-\ln (N)-\ln 2+1-\left(\ln Z_{1}-\ln N+1\right)\right]=0$ hence $\Delta S=0$, which is correct!
5CCP4000 COURSE NOTES ：
\begin{gathered}
P(p) \mathrm{d} p=\frac{1}{N} \quad \frac{V}{h^{3}} 4 \pi p^{2} \mathrm{~d} p \frac{N h^{3}}{V}\left(2 \pi m k_{\mathrm{B}} T\right)^{-\frac{3}{2}} \exp \left(-\frac{p^{2}}{2 m k_{\mathrm{B}} T}\right) \
P(p) \mathrm{d} p=4 \pi p^{2}\left(2 \pi m k_{\mathrm{B}} T\right)^{-\frac{3}{2}} \exp \left(-\frac{p^{2}}{2 m k_{\mathrm{B}} T}\right) \mathrm{d} p
\end{gathered}