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2024 AMC 8

Problem 25

Problem 25 · 2024 AMC 8 Stretch
Counting & Probability complementary-countingcasework

A small airplane has 4 rows of seats with 3 seats in each row. Eight passengers have boarded the plane and are distributed randomly among the seats. A married couple is next to board. What is the probability there will be 2 adjacent seats in the same row for the couple?

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Answer: C — 20/33.
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Hint 1 of 2
Counting WHERE the couple can sit is messy — there are many ways to leave an open pair. Flip it: count the seatings where the couple CAN'T sit together (no open adjacent pair), which is far more rigid, then subtract.
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Hint 2 of 2
The 8 passengers can sit in C(12,8) = 495 ways. Per row L-M-R, an adjacent open pair is blocked exactly when M is taken OR both L and R are taken. Case-split on k = how many of the 4 middle seats are occupied.
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Approach: complementary counting on middle-seat occupancies
  1. Counting the "couple fits" seatings directly is a tangle, so count the COMPLEMENT — seatings with no open adjacent pair anywhere — and subtract from the total. Total ways to seat 8 passengers in 12 seats (order ignored): C(12, 8) = 495.
  2. For NO adjacent pair to be open in a row L–M–R: either M is occupied, or both L and R are occupied. Casework on k = number of rows with M occupied:
  3. k = 0: all four M's empty ⇒ all 8 edge seats filled. 1 way.
  4. k = 1: 4 choices of row, then 2 choices for the extra passenger in that row's edges. 8 ways.
  5. k = 2: C(4,2) = 6 row-choices × C(4,2) = 6 placements of remaining 2 passengers in the 4 unfilled edges. 36 ways.
  6. k = 3: C(4,3) = 4 row-choices × C(6,3) = 20 placements of remaining 3 passengers. 80 ways.
  7. k = 4: all middles filled (4 passengers); C(8,4) = 70 placements of the remaining 4 on edges. 70 ways.
  8. Total "no open pair": 1 + 8 + 36 + 80 + 70 = 195. So the favorable count = 495 − 195 = 300.
  9. Probability = 300495 = 2033. This transfers: when "at least one good spot exists" has many overlapping ways to happen, count the cleaner complement ("no good spot") instead — here the no-pair condition reduced to a tidy per-row rule and a single case-split.
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