I am new to learning combinatorics and I am having trouble with this problem.

There is a round table that rotates with 8 seats. Only 7 people are going to be sitting around this table. 2 of the 7 people are brothers and it is required that there are two people seated in between them (one person and the empty chair is not sufficient). How many ways can they be seated, up to rotation?

I know this is a problem involving permutations, but having that extra empty chair is throwing me off and I cannot wrap my head around this problem. And advice would be great! Thank you!

  • $\begingroup$ If B is Brother, p is person, and O is open, it'll be rotations of BOppBppp, BpOpBppp, BppOBppp, BppBOppp, BppBpOpp, BppBppOp, BppBpppO, which'll be $8*7=56$, but I haven't done this type of math in a while so I'm not positive, and I have no proof $\endgroup$ – Jacob Claassen Sep 8 '17 at 20:55
  • $\begingroup$ exactly 2 or at least 2 ? $\endgroup$ – user451844 Sep 8 '17 at 20:55
  • 3
    $\begingroup$ Arrange them around the table as though there were only seven chairs, keep the eighth chair to the side to be inserted later. Seat the younger brother first and then the older brother and then everyone else. Now, insert the extra chair. $\endgroup$ – JMoravitz Sep 8 '17 at 20:56

Decide the seating order of the people, starting from one of the brothers, say Ivan. Then position the other brother, Alexei, in one of the two slots (fourth and fifth) that fulfill the "separated by two others" condition - $2$ options. Then with Ivan and Alexei resolved, order the remaining five people in one of $5!=120$ ways. Finally add the empty chair to the right of someone, $7$ options, giving $2\cdot 120\cdot 7 = 1680$ options.



  • Brother $A$ has $7$ choices of seats

  • Brother $B$ now has only $2$ choices (one clockwise and one anticlockwise of $A$ )

  • the rest can be permuted in $5!$ ways

  • Thus $7\cdot2\cdot5!\;$ways


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