= Solution
{c}
While the sequence is remembered, the conditional <quantum state> is the known pure endpoint $C|0\rangle$, so its <Bloch vector> is one of the six Pauli vertices. Partial forgetting leaves a probability distribution $p_j$ over possible endpoints, giving
$$
\rho=\sum_{j=1}^6p_j|v_j\rangle\langle v_j|,\qquad
\mathbf s=\sum_jp_j\mathbf v_j.
$$
Hence the vector lies in the <stabilizer octahedron>. The mixing describes lost classical information, not a different unitary evolution of the physical qubit.
Under complete unbiased uncertainty over the six endpoints, $p_j=1/6$, the opposite vectors cancel and
$$
\boxed{\mathbf s=0,\qquad\rho=I/2.}
$$
This is the <maximally mixed state> at the center. Mere loss of the particular sequence does not mathematically force uniform probabilities: a known biased distribution of endpoint states would leave its corresponding noncentral mixture. The central conclusion uses the unbiased randomization implicit in complete forgetting in this setting.
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