91Ó°ÊÓ

(1) For each dot the equation is written as $i\text{as}\left(\frac{1}{V}\right)∫d^3{r}_i$, than for 1 dot multiply it by $N$, for 2 multiply by $N-1$and for 3 dot multiply by $N-2$and so on.

(2) The factor $f_{i{j}_{}}$is written for the line which connects the $i$and $j$dots.

(3) Divide the equation by the symmetric factor of the diagram. Symmetric factor is defined as the permutation of dots that does not alter the figure.

(a) Equation for 2 dots:

$\left(\frac{1}{V^2}\right)∫d^3{r}_1{d}^3{r}_2$

Multiplying the equation with $N,N-1$for 2 dots

$\left(N\right)(N-1)\left(\frac{1}{V^2}\right)∫d^3{r}_1{d}^3{r}_2$

The factor for two dots is$f_{12}$

$\left(N\right)(N-1)\left(\frac{1}{V^2}\right)∫d^3{r}_1{d}^3{r}_2{f}_{12}$

The symmetry factor for the two dots is 2 because the two dots is arranged in two ways.

Thus, the formula for two dots in the figure becomes,

$\left(a\right)=\left(\frac{1}{2}\right)\left(N\right)(N-1)\left(\frac{1}{V^2}\right)∫d^3{r}_1{d}^3{r}_2{f}_{12}$

(b) Equation for 3 dots:

$\left(\frac{1}{V^3}\right)∫d^3{r}_1{d}^3{r}_2{d}^3{r}_3$

Multiplying with $N,N-1,N-2$for 3 dots

$\left(N\right)(N-1)(N-2)\left(\frac{1}{V^3}\right)∫d^3{r}_1{d}^3{r}_2{d}^3{r}_3$

The factor for two dots is $f_{12},f_{13}$

$\left(N\right)(N-1)(N-2)\left(\frac{1}{V^3}\right)∫d^3{r}_1{d}^3{r}_2{d}^3{r}_3{f}_{12}{f}_{13}$

The symmetry factor for the three dots is 2 because the first and third dot is arranged in two ways.

Thus, the formula for three dots in the figure becomes,

$\left(b\right)=\left(\frac{1}{2}\right)\left(N\right)(N-1)(N-2)\left(\frac{1}{V^3}\right)∫d^3{r}_1{d}^3{r}_2{d}^3{r}_3{f}_{12}{f}_{13}$.

(c) Equation for 4 dots:

Multiplying by $N,N-1,N-2,N-3$for 4 dots,

$\left(N\right)(N-1)(N-2)(N-3)\left(\frac{1}{V^4}\right)∫d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4$

The factors for 4 dots are $f_{12}and{f}_{34}$,

$\left(N\right)(N-1)(N-2)(N-3)\left(\frac{1}{V^4}\right)∫d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{f}_{12}{f}_{34}$,

The symmetry factor 4 dots is 8

Thus, the formula for 4 dots become:

$\left(c\right)=\left(\frac{1}{8}\right)\left(N\right)(N-1)(N-2)(N-3)\left(\frac{1}{V^4}\right)∫d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{f}_{12}{f}_{34}$

(d) Equation for 3 dots:

$\left(\frac{1}{V^3}\right)∫d^3{r}_1{d}^3{r}_2{d}^3{r}_3$

Multiplying with $N,N-1,N-2$for three dots,

$\left(N\right)(N-1)(N-2)\left(\frac{1}{V^3}\right)∫d^3{r}_1{d}^3{r}_2{d}^3{r}_3$

The factor for three dots are $f_{12},f_{23},f_{13}$,

$\left(N\right)(N-1)(N-2)\left(\frac{1}{V^3}\right)∫d^3{r}_1{d}^3{r}_2{d}^3{r}_3{f}_{12}{f}_{23}{f}_{13}$

The symmetry factor for 3 dots in the figure is 6, because of the addition of third line in dots

Thus, the formula for the figure is,

$\left(d\right)=\left(\frac{1}{6}\right)\left(N\right)(N-1)(N-2)\left(\frac{1}{V^3}\right)∫d^3{r}_1{d}^3{r}_2{d}^3{r}_3{f}_{12}{f}_{23}{f}_{13}$

(e) Equation for 4 dots:

$\left(\frac{1}{V^4}\right)∫d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4$

Multiplying with $N,N-1,N-2,N-3$for 4 dots,

$\left(N\right)(N-1)(N-2)(N-3)\left(\frac{1}{V^4}\right)∫d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4$

The factor for 4 dots for the figure is,$f_{12},f_{23},\text{and}{f}_{24}$

$\left(N\right)(N-1)(N-2)(N-3)\left(\frac{1}{V^4}\right)∫d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{f}_{12}{f}_{23}{f}_{24}$

The symmetry factor for the figure is 6, attained by putting dot 2 in center and arrange the other dots. So this is arranged in 6 ways.

Thus, the formula for figure (e) is:

$\left(e\right)=\left(\frac{1}{6}\right)\left(N\right)(N-1)(N-2)(N-3)\left(\frac{1}{V^4}\right)∫d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{f}_{12}{f}_{23}{f}_{24}$

(f) Equation for 5 dots:

$\left(\frac{1}{V^5}\right)∫d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{d}^3{r}_5$

Multiplying with $N,N-1,N-2,N-3,N-4$for 5 Dots,

$\left(N\right)(N-1)(N-2)(N-3)(N-4)\left(\frac{1}{V^5}\right)∫d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{d}^3{r}_5$

The factor for 5 dots is $f_{12}{f}_{34}{f}_{45}$.

$\left(N\right)(N-1)(N-2)(N-3)(N-4)\left(\frac{1}{V^5}\right)∫d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{d}^3{r}_5{f}_{12}{f}_{34}{f}_{45}$

The symmetry factor for the figure is 4, by putting dot number 4 in center, the other number is arranged in 4 ways.

Thus, the formula for figure (f) is

$\left(f\right)=\left(\frac{1}{4}\right)\left(N\right)(N-1)(N-2)(N-3)(N-4)\left(\frac{1}{V^5}\right)∫d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{d}^3{r}_5{f}_{12}{f}_{34}{f}_{45}\text{.}$

(g) Equation for 6 dots:

$\left(\frac{1}{V^6}\right)∫d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{d}^3{r}_5{d}^3{r}_6$

Multiplying by $N,N-1,N-2,N-3,N-4,N-5$for 6 dots,

$\left(N\right)(N-1)(N-2)(N-3)(N-4)(N-5)\left(\frac{1}{V^6}\right)∫d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{d}^3{r}_5{d}^3{r}_6$

The factors for 6 dots are $f_{12}{f}_{34}{f}_{56}$,

$\left(N\right)(N-1)(N-2)(N-3)(N-4)(N-5)\left(\frac{1}{V^6}\right)∫d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{d}^3{r}_5{d}^3{r}_6{f}_{12}{f}_{34}{f}_{56}$

The symmetry factor for 6 dots is 48.

Thus, the formula for 6 dots become:

$\left(g\right)=\left(\frac{1}{48}\right)\left(N\right)(N-1)(N-2)(N-3)(N-4)(N-5)\left(\frac{1}{V^6}\right)∫d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{d}^3{r}_5{d}^3{r}_6{f}_{12}{f}_{34}{f}_{56}$

(h) Equation for 4 dots:

$\left(\frac{1}{V^4}\right)∫d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4$

Multiplying by $N,N-1,N-2,N-3$for 4 dots,

$\left(N\right)(N-1)(N-2)(N-3)\left(\frac{1}{V^4}\right)∫d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4$

The factors for figure are $f_{12}{f}_{13}{f}_{24}$,

$\left(N\right)(N-1)(N-2)(N-3)\left(\frac{1}{V^4}\right)∫d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{f}_{12}{f}_{13}{f}_{24}$

The symmetry factor for the figure is 6, attained by putting one dot fixed and arrange the other dots. So this is arranged in 6 ways.

Thus the formula for the figure:

$\left(h\right)=\left(\frac{1}{6}\right)\left(N\right)(N-1)(N-2)(N-3)\left(\frac{1}{V^4}\right)∫d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{f}_{12}{f}_{13}{f}_{24}$