Table 3 Number of combinations of pairs allocations of hamming distance from 0 to r

0:

2ⁿ - 1

1:

(2ⁿ- 1 - n)n + n(n -1) = (2ⁿ -2)n

2:

$\begin{array}{c}\left(2^n-1-\left(\begin{array}{c}\hfill n\hfill \\ \hfill 2\hfill \end{array}\right)\right)\left(\begin{array}{c}\hfill n\hfill \\ \hfill 2\hfill \end{array}\right)+\left(\begin{array}{c}\hfill n\hfill \\ \hfill 2\hfill \end{array}\right)\left(\left(\begin{array}{c}\hfill n\hfill \\ \hfill 2\hfill \end{array}\right)-1\right)=\\ =\left(\begin{array}{c}\hfill n\hfill \\ \hfill 2\hfill \end{array}\right)\left(2^n-2\right)\end{array}$

r:

$\begin{array}{c}\left(2^n-1-\left(\begin{array}{c}\hfill n\hfill \\ \hfill r\hfill \end{array}\right)\right)\left(\begin{array}{c}\hfill n\hfill \\ \hfill r\hfill \end{array}\right)+\left(\begin{array}{c}\hfill n\hfill \\ \hfill r\hfill \end{array}\right)\left(\left(\begin{array}{c}\hfill n\hfill \\ \hfill r\hfill \end{array}\right)-1\right)=\\ =\left(\begin{array}{c}\hfill n\hfill \\ \hfill r\hfill \end{array}\right)\left(2^n-2\right)\end{array}$