I know this is a permutation problem, selecting two from eight. My problem is how to use this information: "must sit one seat away from each other."

In how many ways is it possible to seat eight people at a round table if Alex and Bob must sit one seat away from each other?

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    $\begingroup$ The answer to "must sit exactly one seat away" and "must sit exactly two seats away" will be the same, but "must sit directly opposite" will be different. $\endgroup$ – Henry Jun 2 '16 at 13:04
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    $\begingroup$ It's not selecting two from eight - the question does that by naming Alex and Bob. It's a bit of prework with those two then permuting the remaining six people into the empty chairs. $\endgroup$ – Joffan Jun 2 '16 at 13:07
  • $\begingroup$ I'm not entirely clear whether "one seat away" means Alex and Bob are next to each other, or (my preferred interpretation) there is one intervening guest. Fortunately it makes no difference to the answer, but I'd be interested to know what others think... Something which would affect the answer is if "one seat away" is actually shorthand for "at least one seat away" (e.g. they'll fight if they're next to each other). $\endgroup$ – Joffan Jun 2 '16 at 14:40

Give Alex a choice of the $8$ chairs, then Bob has a choice of $2$ chairs. The remaining guests can arrange themselves in $6!$ ways in the empty chairs left, giving $16\cdot 720 = 11520$ possible arrangements.

  • $\begingroup$ My reading of the question doesn't make it clear whether any one chair is special - or, to put it another way - whether 12345678 is the same as 23456781, or different. You have answered the "different" case. For completeness, the "same" case is one eighth of this (=1440). The direct argument would be that Alex cannot choose a chair because until somebody has sat down, all chairs are indistinguishable - so simply start with "Give Alex a choice of the 1 chair" and continue from there. $\endgroup$ – Martin Kochanski Jun 2 '16 at 13:36
  • $\begingroup$ @MartinKochanski - It nowhere claims that all chairs are indistinguishable, and indeed regardless of the furniture specifications that is entirely unrealistic - for example, in a restaurant, different chairs have different views and different traffic. The round table only ensures that Bob can always have two ways to sit "one seat away" from Alex. $\endgroup$ – Joffan Jun 2 '16 at 13:41


Seat Alex in a random chair. There are now $7$ seats remaining, of which $2$ are next to Alex. If we seat Bob randomly, then with probability $2/7$ he sits next to Alex. So, in $2/7$ of all possible cases, Alex sits next to Bob. There are $8!$ different ways in which you can seat 8 persons, so the number of ways how you can seat $8$ people at a round table so that Alex and Bob sit next to each other is

\begin{equation} \frac{2}{7} \cdot 8! = 11520. \end{equation}

Label the seats $1,2,\ldots,8$. There are $8$ groups of $2$ seats that are next to each other: $\{(1,2),(2,3),\ldots,(7,8),(8,1)\}$. Within each group, there are $2!$ ways to Alex and Bob. Outside each group there are $6!$ ways to seat the remaining $6$ persons. So, the number of ways how you can seat $8$ people at a round table so that Alex and Bob sit next to each other is

\begin{equation} 8 \cdot 2! \cdot 6! = 11520. \end{equation}


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