Solution
Two Round Tables
Show the problem again
Six people are to be seated at two identical round tables. Each table must have at least one person (a single person alone at a table is allowed), and each table can hold all six. Two seatings are the same if every person has the same left-hand neighbour in both (so rotating the people around a table changes nothing, and the tables themselves are indistinguishable), but mirror images are different. A person sitting alone counts as their own neighbour. How many different seatings are there? Give an integer.
Worked solution
**Answer: 274.** A seating is the same thing as a permutation of the six people with exactly two cycles: send each person to their left-hand neighbour, and each table becomes one cycle. So we want the unsigned Stirling number of the first kind c(6, 2). These satisfy c(n, k) = c(n − 1, k − 1) + (n − 1) c(n − 1, k): the newest person either sits alone at a new table, or sits immediately to the right of one of the n − 1 others. For k = 2 this solves to c(n, 2) = (n − 1)! · H_(n−1), where H is the harmonic number. Hence c(6, 2) = 120 × (1 + 1/2 + 1/3 + 1/4 + 1/5) = 120 × 137/60 = 274.
Source: Classic exercise in wide circulation; no traceable original source. Statement written for AxiomIQ.