 Solution

Fix the mode ordering with the $b^{†}$ operator before the $c^{†}$ operator, as in the given state. In the sector with one fermion in each subsystem, use the ordered basisThe state vector is $v=(1/2,1/\sqrt{2},0,1/2{)}^T$, whose squared norm is one. Thus the global density matrix is the rank-one projectorIt has trace one and satisfies ${\rho }_{BC}^2={\rho }_{BC}$, so the global state is pure.

Take the partial trace over $B$. The conditional vectors in $C$ are $v_1=(1/2,1/\sqrt{2}{)}^T$ for $B_1$ and $v_2=(0,1/2{)}^T$ for $B_2$. Therefore the reduced density matrix isIts determinant is $1/16$, giving eigenvaluesBoth are positive; in particular, $\mathrm{Tr}{\rho }_C^2=7/8<1$, so the reduced state is mixed despite global purity. The entanglement entropy is obtained by diagonalizing this reduced matrix and evaluatingFor a globally pure bipartite state this equals the entropy of either reduced subsystem. No numerical evaluation is needed.