= Solution
For the interacting <vacuum state> defined perturbatively by the same <Feynman i-epsilon prescription>, the normalized <time-ordered product> is
$$
\boxed{\left\langle T\prod_{r=1}^m\phi(x_r)\right\rangle
=\left.\frac{1}{Z[\lambda;0]}\frac1{i^m}
\frac{\delta^m Z[\lambda;j]}{\delta j(x_1)\cdots\delta j(x_m)}\right|_{j=0}.}
$$
Each <functional derivative> inserts $i\phi$, accounting for $i^{-m}$. The denominator implements <cancellation of vacuum bubbles>, while disconnected pieces carrying external insertions still contribute. The lower limit $i=i$ in the printed product is a typographical error; the product here runs from 1 to $m$.
At order $V$ in the cubic scalar <field interaction term>, the expansion of the <generating functional> supplies $1/[V!(3!)^V]$. The $V!$ counts permutations of indistinguishable integrated <interaction vertices>, and $3!$ counts permutations of the three field slots at each <interaction vertex>. Applying the <Wick theorem> pairs fields into <Feynman propagators>. Several pairings produce the same <Feynman diagram>; their number $N_G$ multiplies the expansion coefficient. For fixed labelled external points, the net combinatorial weight is
$$
\boxed{w_G=\frac{N_G}{V!(3!)^V}=\frac1{S_G}.}
$$
Here $S_G$ is the order of the graph's <automorphism group> when the external labels are fixed, with permutations of identical parallel lines and reversal of a self-loop included as applicable. This is the <Feynman-diagram symmetry factor> convention in which a <Feynman diagram> is divided by $S_G$. Each cubic <interaction vertex> then contributes $-i\lambda$, each line contributes $\Delta_F$, and each internal vertex is integrated over its position. External labelled points are not divided by $m!$.
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