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Eight distinct people, three of whom are named A, B, and C, are…

Question

Eight distinct people, three of whom are named A, B, and C, are seated in a random order in a row of 8 chairs. Let p be the probability that no two of A, B, C occupy adjacent chairs. Write p = m/n in lowest terms and find m+n.

✓ Verified answer: 19checked by our engine — not a guess

Step-by-step solution

Only the set of positions occupied by {A,B,C} matters for the adjacency condition; the people are interchangeable for counting probability, so compute the probability that a random 3-subset of the 8 positions contains no two consecutive positions.

Total 3-subsets of 8 positions: C(8,3) = 56.
Number of 3-subsets of {1,...,8} with no two consecutive: C(8-3+1, 3) = C(6,3) = 20 (stars-and-bars / gap method).
p = 20/56 = 5/14.
m+n = 5+14 = 19.

Final answer19

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