Problem 11 · 2018 AMC 8
Medium
Counting & Probability
careful-counting
Abby, Bridget, and four of their classmates will be seated in two rows of three for a group picture. If the seating positions are assigned randomly, what is the probability that Abby and Bridget are adjacent to each other in the same row or the same column?
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Answer: C — 7/15.
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Hint 1 of 2
The other four classmates are a distraction. The only thing that matters is which two of the six seats Abby and Bridget land in — so think about seat-pairs, not full seatings.
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Hint 2 of 2
The technique: when an event depends only on the relative position of two items, make the sample space the set of unordered seat-pairs: probability = (adjacent pairs) ÷ (all pairs). Then you just count adjacencies on the grid.
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Approach: count adjacent seat-pairs out of all seat-pairs
- All the randomness boils down to which two of the six seats hold Abby and Bridget: C(6,2) = 15 equally likely unordered pairs.
- Now count the adjacent pairs directly on the 2×3 grid: within a row the 3 seats give 2 side-by-side pairs, × 2 rows = 4 horizontal; each of the 3 columns has one top-bottom pair = 3 vertical. Total 4 + 3 = 7.
- Probability = 7/15 — that's 7/15.
- You'll see it again: reducing "random arrangement" to "random pair of positions for the two people I care about" sidesteps all the irrelevant orderings of everyone else.
Another way — fix Abby, place Bridget:
- Seat Abby first; by symmetry split into where she lands. If Abby takes a middle seat (probability 2/6 = 1/3), it has 3 neighbors among the 5 remaining seats, so Bridget is adjacent with probability 3/5.
- If Abby takes a corner/edge seat (probability 4/6 = 2/3), it has only 2 neighbors, so adjacency probability is 2/5.
- Combine: (1/3)(3/5) + (2/3)(2/5) = 3/15 + 4/15 = 7/15 — same answer, built from Abby's viewpoint.
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