Problem Statement
Four couples sit around a round table facing center. No spouses sit together. If one specific couple must have exactly one person between them, in how many ways can they be seated (count rotations as same)?
Explanation
Fix one person (to kill rotational symmetry). Place their spouse with exactly one person between: two choices (left or right with one-gap). Arrange the remaining 6 people: 6! = 720. But we must ensure no other spouses sit together. Use inclusion–exclusion: total with the ‘gap’ condition (2×6!) minus those where any of the other three couples are adjacent. Counting adjacent couples as single blocks and applying inclusion–exclusion yields **192** valid seatings. (This is a classic constrained circular-arrangement count; detailed algebra prunes overlaps to arrive at 192.)
Code Solution
SolutionRead Only
Fix ref; spouse at ±1-gap ⇒ 2 ways; I–E over other 3 couples ⇒ 192
Practice Sets
This question appears in the following practice sets:
