Question

Keeping only the first two diagrams in equation $8.23$, and approximating $N\approx N-1\approx N-2\approx .....$ expand the exponential in a power series through the third power. Multiply each term out, and show that all the numerical coefficients give precisely the correct symmetry factors for the disconnected diagrams.

Short answer

By doing the expansion, the symmetry factors come out naturally from$Z_c$.

Step-by-step solution

Given Information 

We need to find all the numerical coefficients give precisely the correct symmetry factors for the disconnected diagrams.

Simplify

Lets first write out the expression $8.23$in mathematical form (without diagrams):

$Z_c=\exp \left[\frac{1}{2}\frac{N\left(N-1\right)}{V^2}\int d^3{r}_1{d}^3{r}_2{f}_{12}+\frac{1}{2}\frac{N\left(N-1\right)\left(N-2\right)}{V^3}\int d^3{r}_1{d}^3{r}_2{d}^3{r}_3{f}_{12}{f}_{23}\right]$

Now we can do the approximation $N\approx N-1\approx N-2\dots .$. Then we need to expand 's $Z_c$exp function to the third power:

$$\begin{gathered} Z_c\cong \exp \left[\frac{1}{2}\frac{N^2}{V^2}\int d^3{r}_1{d}^3{r}_2{f}_{12}+\frac{1}{2}\frac{N^3}{V^3}\int d^3{r}_1{d}^3{r}_2{d}^3{r}_3{f}_{12}{f}_{23}\right] \\ \exp \left(\lambda \right)-1+\lambda +\frac{1}{2}{\lambda }^2+\frac{1}{6}{\lambda }^3+\dots \\ {Z}_c=1+\frac{1}{2}\frac{N^2}{V^2}\int d^3{r}_1{d}^3{r}_2{f}_{12}+\frac{1}{2}\frac{N^3}{V^3}\int d^3{r}_1{d}^3{r}_2{d}^3{r}_3{f}_{12}{f}_{23} \\ +\frac{1}{8}\frac{N^4}{V^4}\int d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{f}_{12}{f}_{34}+\frac{1}{8}\frac{N^6}{V^6}\int d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{d}^3{r}_5{d}^3{r}_6\cdot f_{12}{f}_{23}{f}_{45}{f}_{56} \\ +\frac{1}{4}\frac{N^5}{V^5}\int d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{d}^3{r}_5{f}_{12}{f}_{14}{f}_{45}+\dots \end{gathered}$$

To examine those integrals we have to add one more degree of integrals comming from the ${\lambda }^3$of the. This would give:

$$\begin{gathered} Z_c=1+\frac{1}{2}\frac{N^2}{V^2}\int d^3{r}_1{d}^3{r}_2{f}_{12}+\frac{1}{2}\frac{N^3}{V^3}\int d^3{r}_1{d}^3{r}_2{d}^3{r}_3{f}_{12}{f}_{23} \\ +\frac{1}{8}\frac{N^4}{V^4}\int d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{f}_{12}{f}_{34}+\frac{1}{8}\frac{N^6}{V^6}\int d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{d}^3{r}_5{d}^3{r}_6\cdot f_{12}{f}_{23}{f}_{45}{f}_{56} \\ +\frac{1}{4}\frac{N^5}{V^5}\int d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{d}^3{r}_5{f}_{12}{f}_{34}{f}_{45}+ \\ \frac{1}{48}\frac{N^6}{V^6}\int d^3{r}_1{d}^3{r}_2{d}^3{V}_3{d}^3{r}_4{d}^3{r}_5{d}^3{r}_6{f}_{12}{f}_{34}{f}_{56} \\ +\frac{1}{16}\frac{N^7}{V^7}\int d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{d}^3{r}_5{d}^3{r}_6{d}^3{r}_7{f}_{12}{f}_{34}{f}_{56}{f}_{67} \\ +\frac{1}{48}\frac{N^9}{V^9}\int d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{d}^3{r}_5{d}^3{r}_6{d}^3{r}_7{d}^3{r}_8{d}^3{r}_9{f}_{12}{f}_{23}{f}_{45}{f}_{56}{f}_{78}{f}_{89} \end{gathered}$$

We can see that in front of every integral asymmetry factor arises, we can check that on the example of this last integral:

$+\frac{1}{48}\frac{N^9}{V^9}\int d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{d}^3{r}_5{d}^3{r}_6{d}^3{r}_7{d}^3{r}_8{d}^3{r}_9{f}_{12}{f}_{23}{f}_{45}{f}_{56}{f}_{78}{f}_{89}$

If we draw it's diagram and look for dots that can change the place we get exactly $48$permutations:

Explanation 

Another example can be integral:

$\frac{1}{16}\frac{N^8}{V^8}\int d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{d}^3{r}_5{d}^3{r}_6{d}^3{r}_7{d}^3{r}_8{f}_{12}{f}_{34}{f}_{45}{f}_{67}{f}_{78}$

We can see that counting permutations, which is counting the dots with the same role in the integral and possible place where they can be interchanged, is exactly what we get from Taylor's expansion of $\$Z,c\$.$By doing the expansion, the symmetry factors come out naturally.