Find the number of ways in which three couples can be seated around a circular table such that husband and wife are always diametrically opposite to each other? How do I approach this problem? I think the answer should be $2!*2!*2!=8$ but my friend told me that the answer is $3*2!*2!*2!=24$. I think it should be $8$ only. Can anyone help? Please🙏🙏. Thank you


1 Answer 1


Label the people as $M_1, M_2, M_3, W_1, W_2, W_3$, where $M_k$ and $W_k$ must be seated across from each other, where $k \in \{1,2,3\}.$

Then, the problem reduces to determining how many ways there are of seating $M_1, M_2, M_3$, because once they are seated, the seating of $W_1, W_2, W_3$ is then determined.

Since the table is round, without loss of generality, $M_1$ sits in seat-1, which implies that seat-4 must be reserved for $W_1$.

There are therefore $4$ choices for $M_2$. Once $M_1$ and $M_2$ are seated, you now have $2$ seats that are reserved, for $W_1$ and $W_2$.

Therefore, once $M_1$ and $M_2$ are seated there are $4$ seats taken. Therefore, there are then $2$ choices remaining for the seating of $M_3$.

Therefore, the final computation is

$$4 \times 2 = 8 ~~\text{ways}.$$

  • $\begingroup$ But $M_k$ and $W_k$ can switch their seats so wouldn't the answer be $8\times2=16$ ? $\endgroup$
    – Etemon
    Commented Oct 29, 2021 at 20:38
  • $\begingroup$ @Soheil No. Under the assumption that the table is round, seat-1 is reserved for $M_1$. Then, the problem reduces to seating $M_2$ and $M_3$. Any variation in the seating that you might construct by having a married couple switch places has already been counted, as long as you go to the seat where $M_1$ is sitting, label that as seat-1, and then (for example), label the remaining seats in clockwise order (when the table is viewed from above). $\endgroup$ Commented Oct 29, 2021 at 20:41
  • $\begingroup$ @Soheil Proof is in the pudding. Try to construct a seating arrangement that you think that I have overlooked, and then see whether that arrangement is actually part of the $8$ arrangements that I have counted. $\endgroup$ Commented Oct 29, 2021 at 20:43
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    $\begingroup$ If for some reason you insist that rotations are distinct from each other, then you have to multiply by 6 to account for them. I believe your counting scheme already accounts for reflections, so we need not adjust for them. $\endgroup$
    – Kevin
    Commented Oct 30, 2021 at 4:33
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    $\begingroup$ @Kevin Agreed, but in my experience with such math problems, typically, the whole point of specifying circular-table-seating is the unspoken premise that rotations are not distinct from each other. Otherwise, such a problem is generally presented as the people being seated in a row, where each husband-wife pairing must have exactly $2$ other seats between them. $\endgroup$ Commented Oct 30, 2021 at 5:19

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