Probability of 4 couples sitting opposite to each other at a round table with 8 seats

combinatoricspermutationsprobability

After all are seated, I though I can look at four of them. There is a $1/8$ probability of the first person sitting in front of its mate, then $1/7$ for the second, $1/6$ for the third and $1/5$ for the fourth. Yet I'm sure there is still something missing. Should I multiply it by the permutations of each couple? I feel lost at these kinds of questions, I will be glad to hear how does one get to solve them.

Best Answer

Label the chairs$\;1,...,8\;$in clockwise order.

There are $4$ pairs of opposite chairs, namely $(1,5),(2,6),(3,7),(4,8)$.

  • For the first couple (couple #$1$), there are $4$ choices for their pair of opposite chairs, and then $2$ ways for the couple to choose their seats.$\\[4pt]$
  • For the next couple (couple #$2$), there are $3$ choices for their pair of opposite chairs, and then $2$ ways for the couple to choose their seats.$\\[4pt]$
  • For the next couple (couple #$3$), there are $2$ choices for their pair of opposite chairs, and then $2$ ways for the couple to choose their seats.$\\[4pt]$
  • For the last couple (couple #$4$), there is only $1$ choice for their pair of opposite chairs, and then $2$ ways for the couple to choose their seats.$\\[4pt]$

hence there are $$ (4{\,\cdot\,}2) (3{\,\cdot\,}2) (2{\,\cdot\,}2) (1{\,\cdot\,}2) =2^4{\,\cdot\,}4! $$ acceptable seatings out of $8!$ possible seatings.

Thus the probability of an acceptable seating is $$ \frac{2^4{\,\cdot\,}4!}{8!}=\frac{1}{105} $$

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