Ways of arranging of different nationality persons at a round table

$2$ American, $2$ British, $2$ Chinese, $1$ Dutch, $1$ Egyptian, $1$ French and $1$ German people are to be seated for a round table conference, Then

$(a)$ Then number of ways in which no two persons of same nationality are seated together

$(b)$ The number of ways in which only American pair is adjacent, is

$(c)$ The number of ways in which exactly two pairs of same nationality are together, is

$\bf{My\; Try::}$ For $(a)$ Total number of ways of arranging persons in a row is $(10-1)! = 9!$

Now number of ways in which all persons of same nationality sit together is $= 6!\cdot 2!\cdot 2!\cdot 2!$

So number of ways in which all persons of same nationality sit together is $= 9!-6!\cdot 8$

Is my solution for first part right? If not, then how can I solved it? Also, help required in $(b)$ and $(c)$.

Thanks

• @ShreyasS: I’m afraid not: it’s the number of arrangements that have at least one of the pairs separated, but what is wanted is the number that have all three pairs separated. – Brian M. Scott Oct 11 '16 at 10:14
• Yes, I realised that.. deleted the comment – Shreyas S Oct 11 '16 at 10:42

The first problem is an inclusion-exclusion problem. Your $9!-2\cdot8!=357,120$ is the number of arrangements that have at least one of the pairs separated, not the number that have all three pairs separated.

The number of ways in which the two Americans sit together is $2\cdot 8!$: we treat them as a single individual, so we’re seating $9$ individuals, but that one individual has two ‘states’ that have to be distinguished, since the two can sit in either order. Similarly, there are $2\cdot 8!$ arrangements with the two British together and $2\cdot 8!$ with the two Chinese together. Thus, to a first approximation there are $9!-3\cdot2\cdot8!$ arrangements that have no two people of the same nationality seated together.

However, the figure $3\cdot2\cdot8!$ counts twice every arrangement that has both the Americans and the British together. There are $2^2\cdot7!$ such arrangements (why?), so we have to subtract this number from our original approximation. The same goes for the two other pairs, American and Chinese, and British and Chinese: in each of those cases we’ve also counted $2^2\cdot 7!$ arrangements in the figure of $3\cdot2\cdot8!$. Thus, our second approximation is $9!-3\cdot2\cdot8!+3\cdot2^2\cdot7!$.

Unfortunately, this still isn’t quite right: the $2^3\cdot6!$ arrangements that have the two Americans together, the two British together, and the two Chinese together were counted once each in the $9!$ term; subtracted $3$ times each in the $-3\cdot2\cdot8!$ term; and added back in $3$ times in the $3\cdot2^2\cdot7!$ term, so they have been counted a net total of one time each. But we don’t want to count them, so we have to subtract $1$ for each of them to reach the final answer:

\begin{align*} 9!-3\cdot2\cdot8!+3\cdot2^2\cdot7!-2^3\cdot6!&=(9-6)\cdot8!+(84-8)\cdot6!\\ &=3\cdot8!+76\cdot6!\\ &=244\cdot720\\ &=175,680\;. \end{align*}

To count the arrangements that have just the two Americans together, start with the $2\cdot8!$ arrangements that have them together. Among these there are $2^2\cdot7!$ that also have the two British together (why?), and $2^2\cdot7!$ that also have the two Chinese together. Finally, there are (as we saw before) $2^3\cdot6!$ arrangements that have all three pairs together. The inclusion-exclusion calculation is simpler this time:

$$2\cdot8!-2\cdot2^2\cdot7!+2^3\cdot6!=46,080\;.$$

I’ll leave the last one for you to try now that you’ve seen the first two.