Seating three couples with no pair together

Question

Three couples, six people total, sit in a random order along a row of 6 seats. Every ordering is equally likely. What is the probability that no couple sits in two adjacent seats?

Accepted Answer

Total orderings: 6! = 720. Count the "bad" orderings where at least one couple is adjacent, using inclusion–exclusion. Let Ai be the event that couple i sits together. One couple together. Glue that couple into a block (2 internal orders), leaving 5 items to arrange: Ai = 2	imes 5! = 240,\qquad 	ext{three couples} \Rightarrow 3	imes 240 = 720. Two couples together. Two blocks plus 2 singles, 4 items, each block has 2 internal orders: Ai\cap Aj = 2^2	imes 4! = 96,\qquad \binom{3}{2}=3 \Rightarrow 3	imes 96 = 288. All three together. Three blocks, 3 items: A1\cap A2\cap A3 = 2^3	imes 3! =…

Seating three couples with no pair together

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