Fourteen distinct persons are seated randomly around a circular table. What is the probability that three particular persons always sit together as neighbours, in any order, while the remaining persons occupy the other seats?

Introduction / Context:
This question combines circular permutations with basic probability. We are given 14 distinct persons seated around a round table and we want the probability that three particular persons always sit together as neighbours. In circular arrangements, rotating everyone together does not create a new arrangement, so we must use the idea of circular permutations. The key concept is to treat the three special persons as a single block and then count how many such favourable arrangements exist compared with the total possible arrangements of all persons.

Given Data / Assumptions:

There are 14 distinct persons in total.

Three of these persons are special and must sit together as neighbours.

The table is circular, so arrangements that differ only by rotation are considered the same.

All circular seating arrangements are equally likely.

Concept / Approach:
For n distinct persons seated around a circular table, the number of possible arrangements is (n - 1)!. This is because we can fix one person as a reference to break circular symmetry and then arrange the remaining persons linearly. For the favourable cases, we treat the three special persons as one combined block and count circular arrangements of this block plus the remaining persons. We also multiply by the number of internal permutations of the three persons within that block.

Step-by-Step Solution:
Step 1: Total number of circular arrangements of 14 distinct persons = (14 - 1)! = 13!.Step 2: For favourable arrangements, treat the three particular persons as one single block.Step 3: Then we effectively have 1 block + 11 other persons = 12 objects to arrange around the table.Step 4: The number of circular arrangements of these 12 objects = (12 - 1)! = 11!.Step 5: Inside the block, the three particular persons can be arranged among themselves in 3! = 6 ways.Step 6: Therefore, number of favourable arrangements = 11! * 3!.Step 7: Probability = favourable / total = (11! * 3!) / 13!.Step 8: Simplify 13! as 13 * 12 * 11!, so probability = 6 / (13 * 12) = 6 / 156 = 1 / 26.

Verification / Alternative check:
We could instead fix one of the three special persons as a reference and count ways to place the other two next to that person, but after taking into account the remaining persons and circular symmetry, we again obtain 1 / 26. Any correct method must yield the same simplified fraction, so the value 1 / 26 is consistent.

Why Other Options Are Wrong:
3/26 and 5/26 correspond to larger favourable sets than are actually possible and would require more arrangements than exist. The fraction 1/13 doubles the correct probability, treating linear permutations instead of circular ones. The fraction 2/13 still overestimates the chance that three specific persons stay together and does not match the correct ratio of favourable to total circular arrangements.

Common Pitfalls:
A common mistake is to forget that the table is circular and to count all 14! linear arrangements, which overcounts each circular arrangement 14 times. Another error is to treat the three special persons as fixed in one spot rather than as a movable block, which distorts the count of favourable cases. Always remember to use (n - 1)! for circular arrangements and to multiply by internal permutations when treating several people as a single block.

Final Answer:
The required probability that the three particular persons always sit together is 1/26.