In how many ways can two Americans, two British, one Chinese, one Dutch and one Egyptian sit at a round table so that persons of the same nationality are always separated? (Clockwise and anti-clockwise arrangements are different.)

My attempt:
I first arranged the Dutch, Egyptian and Chinese gentlemen on the round table in such a way that there is one empty chair between any two persons and two empty chairs between a single pair. This can be done in 2! ways. Then I choose a Britishman and an American each as well as two seats for them and arrange them in 2! ways. Now the remaining pair of an American and a Britishman have to sit together and the no. of internal arrangements for then would also be equal to 2!. Multiplying all of these gives me answer 96. But my book mentions the answer as 336. I think my method is incorrect. Can anyone suggest me the correct method?

  • 1
    $\begingroup$ I think your first move (empty spots between D,E,C) is not appropriate. E.g. how about DECABAB (as a circle of course)? $\endgroup$
    – drhab
    Oct 27 '16 at 9:40

Fix the Egyptian. Then we can use the inclusion–exclusion principle to find the number of arrangements:

  • There are $6!=720$ ways to seat the other people, ignoring restrictions.
  • Of these, $5!\cdot2=240$ ways seat the Americans together and are excluded (this may be seen by treating the Americans as a block; note that they can sit in two ways internally). The same number applies for the British.
  • $4!\cdot2\cdot2=96$ ways seat both Americans and both British together, and are included.

So there are $720-240-240+96=336$ ways to seat the people with the given restrictions.


$\underline{Another\; way}\quad$

We denote the people as $A,a,B,b,C,D,E,\;$ depict arrangements linearly with $E$ at the ends,
and apply the "gap" method followed by the "subtraction" method

Firstly, we shall keep the Americans separate by placing them in the gaps of $E-B-b-C-D-[E]$ and permute the other letters, thus $(5\cdot4)4! = 480$

We shall now subtract arrangements with the Britishers together treating them as a super $B$ which can be internally permuted in $2$ ways,
$ E-\mathscr B - C - D - [E]\;\; so \;\; (4\cdot3)3!\times2 = 144$

thus the final ans is $480 -144 = 336$

  • $\begingroup$ I didnt quite understand the second part of your solution.... $\endgroup$
    – SirXYZ
    Oct 27 '16 at 15:26
  • $\begingroup$ Some of the arrangements in the first part would have $B$ and $b$ together as $\boxed{Bb}\; or\;\boxed{bB}$. The second part counts such arrangements, and subtracts them. The Americans can now only be placed in $4\cdot3$ ways, and the others in $3!\times 2$ ways, the final $2$ being for $\boxed{Bb}$ and $\boxed{bB}$. $\endgroup$ Oct 27 '16 at 15:30
  • $\begingroup$ Yeh....I got that logic but can you explain how you obtained the nos. to calculate the arrangements to be subtracted... $\endgroup$
    – SirXYZ
    Oct 27 '16 at 15:36
  • $\begingroup$ Have you read my comment ? $\endgroup$ Oct 27 '16 at 15:38
  • $\begingroup$ What I am asking is that how you got the (4.3)×3!.....I understood that this should be multiplied by 2 for internal arrangements of the British. $\endgroup$
    – SirXYZ
    Oct 27 '16 at 15:40

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