Wick theorem exercise I'm a newbie in QFT and I have some doubts with this simple exercise:
Using the Wick Theorem evaluate
$$\langle0|T(\phi^4(x)\phi^4(y)|0\rangle$$

*

*Use a diagrammatic approach to represent the possible contractions
(how many $\phi$-lines are attached to each vertex).


*Determing numerical factor in front of each diagrams.
 A: First of all the Wick's theorem states
$$
T(A B C \ldots Y Z)=:\{A B C \ldots Y Z \,+"\text { all contractions" }\}:
$$
(In the case of fermions we have to take care about anticommutation relations,
i.e. every time when we interchange neighboring fermionic operators a minus sign appears.)
In this exercise you shall apply Wick's theorem for bosons.
It is clear that all normal-ordered terms fall off, because their vacuum expectation value is equal to zero. Thus the only remaining terms are those with four contractions.
If you contract one $\phi(x)$ with one $\phi(y)$ four times you shall get $(\langle 0|T(\phi(x) \phi(y))| 0\rangle)^{4} .$ This can be done in $4 !=24$ ways. The next possibility is to make two contractions between fields $\phi(x)$ and $\phi(y) .$ One field $\phi(x)$ can be contracted in 4 ways with one of the $\phi(y)^{\prime}$ s. The next $\phi(x)$ can be contracted in three ways with one of the remaining $\phi(y)^{\prime} \mathrm{s} \cdot$ The obtained result has to be multiplied by $6,$ because this is the number of ways in which two fields $\phi(x)$ can be chosen from the four possible. Thus, there are $4 \cdot 3 \cdot 6=72$ possible contractions of this type. There are three mutual contractions between two fields $\phi(x),$ the similar is obtained for fields $\phi(y),$ so the corresponding coefficient is 9. Thus,
\begin{equation}
\begin{aligned}
\left\langle 0\left|T\left(\phi^{4}(x) \phi^{4}(y)\right)\right| 0\right\rangle &= 24 \,\langle 0|T(\phi(x) \phi(y))| 0\rangle^{4}\\\\
&+72\,\langle 0|T(\phi(x) \phi(x))| 0\rangle\langle 0|T(\phi(y) \phi(y))| 0\rangle(\langle 0|T(\phi(x) \phi(y))| 0\rangle)^{2} \\\\
&+9\,(\langle 0|T(\phi(x) \phi(x))| 0\rangle)^{2}(\langle 0|T(\phi(y) \phi(y))| 0\rangle)^{2} \\\\
&=24\,\left(\mathrm{i} \Delta_{\mathrm{F}}(x-y)\right)^{4}+72\left(\mathrm{i} \Delta_{\mathrm{F}}(x-x)\right) \mathrm{i} \Delta_{\mathrm{F}}(y-y)\left(\mathrm{i} \Delta_{\mathrm{F}}(x-y)\right)^{2} \\\\
&+9\,\left(\mathrm{i} \Delta_{\mathrm{F}}(x-x)\right)^{2}\left(\mathrm{i} \Delta_{\mathrm{F}}(y-y)\right)^{2}
\end{aligned}
\end{equation}
where $\Delta_{\mathrm{F}}(x-y)$ is the Feynman propagator.
The last expression can of course represented by Feynman diagrams.
Now, go through these steps and depict the last expression as diagrams. Good luck and if you have some trouble with the computation/drawing feel free to ask for details.
