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It is the last problem of the AHSME competition 1988-1989 (question 30)

"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 $\text{GBBGGGBGBGGGBGBGGBGG}$ we have that $S=12$. The average value of $S$ (if all possible orders of these $20$ people are considered) is closest to

$\text{(A)}\ 9\qquad\text{(B)}\ 10\qquad\text{(C)}\ 11\qquad\text{(D)}\ 12\qquad\text{(E)}\ 13 $

(see this link for the source: http://www.artofproblemsolving.com/Wiki/index.php/1989_AHSME_Problems/Problem_30)

I can't find an exact solution, but i know the answer must be A:9 (after having programmed it the exact solution is 91/10)

I'm in the sixth year of secondary school if that gives an idea of my mathematics level. (and i've completed all the AHSME questions from 1985-1986 to 1994-1995 unless this one)

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Given a fixed $n\in\{1,2,3,...,19\}$, what is the probability that the kids in position $n$ and $n+1$ are different sex?

That is easily computed as $\frac{7\cdot 13+13\cdot7}{20\cdot 19}$.

So the expected number of "boy/girl" or "girl/boy" pairs in position $n$ is $\frac{91}{10\cdot 19}$.

But the expected number of boy/girl pairs across the whole line is just the sum over each $n$. Since there are $19$ values of $n$, it is $19\cdot\frac{91}{10\cdot19}=\frac{91}{10}$.

That you can do this sort of summing of expected values is a bit confusing when the value in each location is not independent. But you can.

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    @ThomasAndrews: What a superb solution. Thank you!2017-03-23