If 15 people are seated randomly around a circular table, what is the probability that three specific friends will be seated next to each other?

Instruction: Seat 15 distinct people uniformly around a circular table. Find the probability that three specified friends form one consecutive three-seat block, in any internal order, allowing wraparound.

Context: This question assesses the candidate's ability to solve a circular permutation problem and understand the nuances of rotational symmetry in probability.

Updated

Official answer

Interpret the event as the three specified friends occupying three consecutive seats, in any internal order, including across an arbitrarily chosen start of the circle. All 15 distinct people are seated uniformly at random.

Up to rotation, there are 14! unrestricted arrangements. Treating the three friends as one block leaves 13 objects, with 12! circular arrangements. The friends have 3! internal orders, so

P = 12!×3!/14! = 6/(14×13) = 3/91, about 3.30%.

Using 13! for the block arrangement would handle rotations inconsistently with the denominator. An alternative check uses the friends' three-seat subset: 15 of the C(15,3)=455 subsets are consecutive, giving 15/455 = 3/91. The event does not require each friend to be adjacent to both others.

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