1989 AHSME Problems/Problem 30

Problem

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

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 $\boxed{9\tfrac1{10}(A)}$ . The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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1989 AHSME Problems/Problem 30

Problem

Solution