= Solution
\b[Yes: deterministic entanglement swapping gives the required singlet.] Denote Bob's two local qubits by $B_1,B_2$. Known <local unitary operations> first convert the two given <spin singlet states> into $|\Phi^+\rangle_{AB_1}$ and $|\Phi^+\rangle_{B_2C}$. For example applying $ZX$ to the second qubit of a singlet gives $|\Phi^+\rangle$, where $X,Z$ are <Pauli operators>.
Use the binary-labelled <Bell states>
$$
|B_{pq}\rangle=(I\otimes X^qZ^p)|\Phi^+\rangle,\qquad p,q\in\{0,1\}.
$$
They are $\Phi^+,\Phi^-,\Psi^+,\Psi^-$ for labels $(0,0),(1,0),(0,1),(1,1)$. Reordering registers to group Bob's qubits, direct expansion gives
$$
|\Phi^+\rangle_{AB_1}|\Phi^+\rangle_{B_2C}
=\frac12\sum_{p,q=0}^1|B_{pq}\rangle_{B_1B_2}|B_{pq}\rangle_{AC}.
$$
All four basis vectors are real in this convention; otherwise the paired expansion must account for complex conjugation. Bob performs the <Bell-basis measurement> on $B_1B_2$. This joint operation is local to him, so it is allowed by <LOCC>. Each outcome occurs with probability $1/4$ and leaves $AC$ in the corresponding <Bell state>.
Bob sends $(p,q)$ to Charlie. Charlie applies $Z^pX^{-q}$ to obtain $|\Phi^+\rangle_{AC}$, and then $XZ$ to obtain
$$
\boxed{|\Psi^-\rangle_{AC}=\frac{|01\rangle-|10\rangle}{\sqrt2}.}
$$
The correction works for every outcome, so no postselection or successful-outcome assumption is needed. This is <entanglement swapping>, equivalently teleportation of $B_1$ to $C$ while preserving its entanglement with $A$. If Bob's measurement result is ignored, the average $AC$ state is $I_4/4$; the protocol therefore does not supply entanglement usable without the classical outcome or violate the <no-communication theorem>.
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