= Solution
For ordinary bosonic operators at distinct times, <time ordering> places the later operator to the left:
$$
\boxed{T\{O_1(t_1)O_2(t_2)\}=\begin{cases}O_1(t_1)O_2(t_2),&t_1>t_2,\\O_2(t_2)O_1(t_1),&t_2>t_1.\end{cases}}
$$
Equivalently, multiply these two products by the corresponding <Heaviside step functions>. For two odd fermionic operators the second product receives a minus sign; more generally a graded interchange has sign $(-1)^{F_1F_2}$. This sign is the <fermionic sign> convention for time-ordered field products, not a minus sign for arbitrary bosonic operators. No equal-time convention is needed for the two distinct times in this clause.
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