Suppose that $7$ boys and $13$ girls line up in a row. Let $S$ be the number of places in the row where a boy and a girl are standing next to each other. For example, for the row $GBBGGGBGBGGGBGBGGBGG$ we have $S=12$. The average value of $S$ (if all possible orders of the 20 people are considered) is closest to

$ \textbf{(A)}\ 9 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 11 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 13 $
1. **Define the problem and use linearity of expectation:**
   We need to find the average value of \( S \), the number of places where a boy and a girl are standing next to each other in a row of 7 boys and 13 girls. By the linearity of expectation, we can calculate the expected value of \( S \) by finding the probability that any two consecutive positions are occupied by a boy and a girl, and then multiply this probability by the number of consecutive positions (which is 19).

2. **Calculate the probability of a boy and a girl standing next to each other:**
   There are \( \binom{20}{2} = \frac{20 \times 19}{2} = 190 \) possible ways to choose any two consecutive positions out of 20.
   
   The number of ways to choose 1 boy out of 7 and 1 girl out of 13 is \( \binom{7}{1} \times \binom{13}{1} = 7 \times 13 = 91 \).

   Therefore, the probability that any two consecutive positions are occupied by a boy and a girl is:
   \[
   \frac{91}{190}
   \]

3. **Calculate the expected value of \( S \):**
   Since there are 19 consecutive positions in a row of 20 people, the expected value of \( S \) is:
   \[
   19 \times \frac{91}{190} = 19 \times \frac{91}{190} = 19 \times 0.478947368 = 9.1
   \]

4. **Round to the nearest integer:**
   The expected value of \( S \) is approximately 9.1, which is closest to 9.

Conclusion:
\[
\boxed{9}
\]