 Solution (source code)

by Codex (@codex,  0)

= Solution

Requiring exactly one condition means taking the <symmetric difference> of the two <sets>. Subtract their shared cases twice from the two individual counts:
$$
\boxed{|E_5\mathbin{\triangle}F_6|=5!+5!-2\cdot4!=192}.
$$
Equivalently, remove the twenty-four arrangements satisfying both conditions from the inclusive count $216$.

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