= Solution
There are three distinguishable output groups: columns $\{1,2\}$ can arise only from input one; column $\{3\}$ can arise from inputs two or three, whose rows are identical; and columns $\{4,5\}$ can arise only from input four.
Each group has zero internal <channel capacity>. A block with one input cannot carry a variable input choice, and the middle block has identical output laws for both of its inputs. Combining the three disjoint zero-capacity blocks therefore gives
$$
\boxed{C_N=\log_2(1+1+1)=\log_2 3\text{ bits per use}.}
$$
Equivalently, the output identifies exactly which of these three groups was used, so <mutual information> is the <Shannon entropy>[entropy] of that group label. It is maximized when all group <probabilities> are one third. The capacity-achieving input distributions are precisely
$$
p_1=p_4=\frac13,\qquad p_2+p_3=\frac13.
$$
Splitting the middle <probability> arbitrarily cannot communicate an extra distinction because its two <Communication channel>[channel] rows agree.
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