My notes introduce the truncated expectation in the following way: given $S_0$ a quadratic form, consider a generating function $$e^{W(J)} = \int \prod_x d\varphi_x e^{-S_0+ JO} = \int P(d\varphi) e^{JO}. $$ Let define the truncated expectation of $O$ as $$ E^T(O;n) = \frac{\partial^n}{\partial J^n} W(J) \bigg|_{J=0}= \frac{\partial^n}{\partial J^n} \log \int P(d\varphi) e^{JO} \bigg|_{J=0} $$ And so for example $E^T(O;1)=\frac{\partial}{\partial J} \log \int P(d\varphi) e^{JO} \bigg|_{J=0} =\frac{\int P(d\varphi) O}{\int P(d\varphi) }=\langle O\rangle $.
Now, using the Wick rule, I calculate $$E^T(\varphi_1 \varphi_2 \varphi_3 \varphi_4;1)=\langle \varphi_1 \varphi_2 \varphi_3 \varphi_4\rangle = \langle \varphi_1 \varphi_2\rangle \langle \varphi_3 \varphi_4\rangle +\langle \varphi_1 \varphi_4\rangle \langle \varphi_2 \varphi_3\rangle +\langle \varphi_1 \varphi_3\rangle \langle \varphi_2 \varphi_4\rangle. $$
The rules for Feynman diagrams are: construct a Feynman diagram drawing a segment to each field and then join the different segments if there is a contraction between the field, that is if I have the expectation between the field. Then, following this rule I have that $\langle \varphi_1 \varphi_2 \varphi_3 \varphi_4\rangle $ is composed by disconnected diagram, because, for example for the first term that comes from the Wick rule, $\langle \varphi_1 \varphi_2\rangle \langle \varphi_3 \varphi_4\rangle $ I have one segment that connects the vertices $1-2$ and one segment for $3-4$.
Where am I wrong?