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2011 AMC 8

Problem 12

Problem 12 · 2011 AMC 8 Easy
Counting & Probability fix-one-position

Angie, Bridget, Carlos, and Diego are seated at random around a square table, one person to a side. What is the probability that Angie and Carlos are seated opposite each other?

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Answer: B — 1/3.
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Hint 1 of 2
Don't track all four people — only the relationship between Angie and Carlos matters. Plant Angie in a seat and ask: where does Carlos land?
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Hint 2 of 2
Once Angie is fixed, Carlos is equally likely in each of the 3 remaining seats. The question becomes "how many of those 3 are opposite her?"
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Approach: fix one person as a reference, then place the other
  1. Pin Angie in any one seat — this throws away the seating's overall symmetry so we only watch Carlos.
  2. Carlos is equally likely to take any of the 3 leftover seats, and exactly one of them is directly across from Angie.
  3. So the probability is 1 out of 3 = 1/3.
  4. Why this transfers: in seating/circular problems, fixing one object as an anchor turns a messy 4! = 24 count into a simple "where does the one I care about go?"
Another way — count all arrangements:
  1. Total ways to seat 4 people = 4! = 24.
  2. Favorable: choose Angie's seat (4 ways), Carlos takes the opposite seat (1 way), the other two fill the rest (2! = 2). That's 4 × 1 × 2 = 8.
  3. Probability = 8/24 = 1/3.
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