Relation between symmetry factors

In $\phi^3$ theory, the generating functional for interacting field theory is given by: $$Z_1(J) = \sum_{V=0}^{\infty} \frac{1}{V!} \Big[ \frac{iZ_g g}{6} \int \Big( \frac{1}{i}\frac{\delta}{\delta J}\Big)^3 d^4 x \Big]^V \times \sum_{P=0}^{\infty} \frac{1}{P!} \Big[ \frac{i}{2} \int J(y) \Delta(y-z) J(z) \, d^4 y \, d^4z \Big]^P$$

[Reference: Srednicki: eqn. (9.11)]

Let, for specific values of $V$ and $P$ we get some terms from it. One of them is a disconnected diagram consisted of two connected diagrams $C_1$ and $C_2$. The disconnected diagrams symmetry factor is, say, $S$; that is the term for disconnected diagram has a numerical coefficient: $\frac{1}{S}$. Now we write the term for disconnected diagram according to the eqn (9.12): $$D = \frac{1}{S_D} \prod_I (C_I)^{n_I}$$

where $n_I$ is an integer that counts the number of $C_I$ ’s in $D$, and $S_D$ is the additional symmetry factor for $D$. Here, $S_D = \prod_I n_I !$

In this case is this true: $S=\frac{1}{n_1!} \times \frac{1}{n_2!} \times C_1$'s symmetry factor $\times C_2$'s symmetry factor?

Yes, it is true. For $V=2$, $P=4$ and $E = 2$ it can be proved.