1989 AMC 12 Problem 30

Attempt Problem 30 of the 1989 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1989 AMC 12 solutions, or check the answer key.

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30.

Suppose that 77 boys and 1313 girls line up in a row. Let SS 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 GBBGGGBGBGGBBGGGBGBGGGBGBGGBGGGGBGBGGBGG we have S=12.S=12. The average value of SS (if all possible orders of these 2020 people are considered) is closest to
99
1010
1111
1212
1313
Answer: A
Concepts:expected valuedouble countingpermutations
Small hint:
Use an indicator for each of the 1919 adjacent pairs
Big hint:
For a fixed adjacent pair, compute the probability of seeing BGBG or GBGB
Solution:
For each of the 1919 adjacent position pairs, the probability of mixed sexes is 720⋅1319+1320⋅719. \frac7{20}\cdot\frac{13}{19} +\frac{13}{20}\cdot\frac7{19}. By linearity of expectation, E[S]=19⋅2⋅7⋅1320⋅19=9110=9.1, \begin{aligned} \mathbb E[S] &=19\cdot\frac{2\cdot7\cdot13}{20\cdot19}\\ &=\frac{91}{10}=9.1, \end{aligned} which is closest to 9.9.
Thus the correct answer is A.
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