Doubt in the solution of a competition combinatorial problem concerning the average number of times the letters of a sequence change I am currently going through  Andreescu's 102 combinatorial problems. I couldn't solve number 28 after repeated attempts, so I checked the solution. The problem and solution go as follows:

However, I do not understand why summing the $N_i$'s and dividing over $20!$ is the average. As I see it, if $S_i$ denotes the number of rows with $i$ boy-girl or girl-boy appearances in the row, the average should be
$\dfrac{\sum_{i=2}^{13}iS_i}{\binom{20}{7}}$.
I was hoping that someone would throw some light on why my formula is incorrect and the logic behind the solution (I am relatively new to combinatorial problems in competitions.)

Sometimes it is helpful to reduce the problem to smaller numbers so that it can be calculated manually. Here we take $5$ instead of $20$ persons and consider a group of two boys and three girls.
\begin{align*}
\{B_1,B_2,G_1,G_2,G_3\}\tag{1}
\end{align*}
In order to analyse the situation it is sufficient to consider the
\begin{align*}
\color{blue}{\binom{5}{2}=10}
\end{align*}
different rows
\begin{align*}
\begin{array}{l|c}
B\quad B\color{blue}{\bullet} G\quad G\quad G&1\\
B\color{blue}{\bullet} G\color{blue}{\bullet} B\color{blue}{\bullet} G\quad G&3\\
B\color{blue}{\bullet} G\quad G\color{blue}{\bullet} B\color{blue}{\bullet} G&3\\
B\color{blue}{\bullet} G\quad G\quad G\color{blue}{\bullet} B&2\\
G\color{blue}{\bullet} B\quad B\color{blue}{\bullet} G\quad G&2\\
G\color{blue}{\bullet} B\color{blue}{\bullet} G\color{blue}{\bullet} B\color{blue}{\bullet} G&4\\
G\color{blue}{\bullet} B\color{blue}{\bullet} G\quad G\color{blue}{\bullet} B&3\\
G\quad G\color{blue}{\bullet} B\quad B\color{blue}{\bullet} G&2\\
G\quad G\color{blue}{\bullet} B\color{blue}{\bullet} G\color{blue}{\bullet} B&3\\
G\quad G\quad G\color{blue}{\bullet} B\quad B&1
\end{array}
\end{align*}
Each of these rows represents the $2!\cdot 3!=12$ pairwise different configurations when we use the elements from (1) giving a total of $12\cdot 10=5!$ rows. But this is not so important, as we can only focus on $BG$ and $GB$ combinations which are marked with $\color{blue}{\mathrm{blue}}$ dots.
One approach:
We get the multiplicities from the right-most column of the table above and derive according to OPs approach
\begin{align*}
\color{blue}{\frac{1\cdot S_1+2\cdot S_2 + 3\cdot S_3+4\cdot S_4}{\binom{5}{2}}}=\frac{1\cdot 2+2\cdot 3+3\cdot 4+4\cdot 1}{10}
\color{blue}{=2,4}
\end{align*}
Another approach:
We focus on the combinations $BG$ and $GB$ for each of the four position pairs $(1,2), (2,3), (3,4), (4,5)$.
\begin{align*}
\begin{array}{ccccc}
B&G&\cdot&\cdot&\cdot\\
\cdot&B&G&\cdot&\cdot\\
\cdot&\cdot&B&G&\cdot\\
\cdot&\cdot&\cdot&B&G\\
G&B&\cdot&\cdot&\cdot\\
\cdot&G&B&\cdot&\cdot\\
\cdot&\cdot&G&B&\cdot\\
\cdot&\cdot&\cdot&G&B\\
\end{array}
\end{align*}
We fix a specific position $j, 1\leq j\leq 4$ and let's say the combination $BG$. We calculate the probability of the event that $BG$ occurs at a specific position $j$ by dividing the favorable number of possibilities by the total number which is as we have two boys and three girls:
\begin{align*}
\color{blue}{\frac{2\cdot 3}{5\cdot 4}}\tag{2}
\end{align*}
We note that (2) is strongly related with $N_j$ from OPs problem, since we can (2) write as
\begin{align*}
\frac{2\cdot 3}{5\cdot 4}=\frac{2\cdot 3\cdot 3!}{5\cdot 4\cdot 3!}=\frac{2\cdot 3\cdot 3!}{5!}\tag{3}
\end{align*}
Doubling (3) since we have to respect $BG$ as well as $GB$ at a specific position and using the linearity of expectation we calculate according to (2) and (3)
\begin{align*}
\color{blue}{\frac{N_1+N_2+N_3+N_4}{5!}}=4\cdot \frac{2 (2\cdot 3) 3!}{5!}=4\cdot\frac{ 2(2\cdot 3)}{5\cdot 4}\color{blue}{=2,4}
\end{align*}
and both results coincide.
