1989 AHSME Problems/Problem 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

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

Solution 1

We approach this problem using Linearity of Expectation. Consider a pair of two people standing next to each other. Ignoring all other people, the probability that a boy is standing on the left position and a girl is standing on the right position is $\frac7{20}\cdot\frac{13}{19}$. Similarly, if a girl is standing on the left position and a boy is standing on the right position the probability is also $\frac{7\cdot 13}{20\cdot 19}$. Thus, the total probability of the two people being one boy and one girl is $\frac{91}{190}$.

There are a total of 19 different adjacent pairs, so by Linearity of Expectation, we have that the expected value of $S$ is $\frac{91}{10} \to \boxed{A}$.

Solution 2

Suppose that the class tried every configuration. Boy $i$ and girl $j$ would stand next to each other in $2$ different orders, in $19$ different positions, $18!$ times each. Summing over all $i,j$ gives $7\cdot13\cdot2\cdot19\cdot18!=\tfrac{91}{10}\cdot20!$, so the average value of $S$ is $\boxed{9\tfrac1{10}(A)}$.

See also

1989 AHSME (ProblemsAnswer KeyResources)
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