Number of arrangements in a rectangular table (II)

I guess I found the supposed solution. Firstly a rectangular table is not a round table. I suppose all diplomats should sit at the same side, so the table is a line. Secondly if we reverse the order we should count it as the same arrangement. (I. e. two arrangements are distinct if at least one diplomat has different set of neighbors.)

Let's consider Chinese delegation as a single metadiplomat and the same for Japanese delegation. Then we have $13$ diplomats, $2$ of which can't sit together. There are $13!$ arrangements in total and $2 \cdot 12!$ of them contain Chinese and Japanese metadiplomats next to each other. So we have $13! - 2 \cdot 12!$ arrangements.

Now remember that Chinese delegation can sit in $3!$ ways and Japanese delegation can sit in $4!$ ways within their places. And reversing order gives the same arrangements. Then we have

$$\frac{(13! - 2 \cdot 12!) \cdot 3! \cdot 4!}{2} = 379369267200$$

arrangements.

P. S. For a round table under the same assumption about distinct arrangements we start with Chinese metadiplomat, have $10$ (of $12$) places for Japanese metadiplomat and $11!$ ways to arrange other diplomats. Taking into account possible reverse of order we get

$$\frac{10 \cdot 11! \cdot 3! \cdot 4!}{2} = 28740096000$$

arrangements, that is $11$ times less than you answer.

If two arrangements are distinct if at least one diplomat has another seat then we should not divide by $2$, but should multiply by $18$ possible options for the leftmost diplomat of Chinese delegation and get $$10 \cdot 11! \cdot 3! \cdot 4! \cdot 18 = 1034643456000$$ arrangements.

In your reasoning you don't have $11$ places for $11$ diplomats, so you should count only $10!$ cyclic orders, but not $11!$. Then you will get $28740096000$.

NEW

I've noticed we could have $J$ at one end, and $C$ at the other, which is also an illegal arrangement.

So I tried this instead, which gives an answer of $465,589,555,200$, but seems logically sound.

Label the chairs $1\dots18$. Note that starting from $1$ is the same as starting from $10$. First place the Chinese trio ($18$ ways times $6$ for internal positions), and then the Japanese ($9\times24$), and then the $11$ remaining diplomats ($11!$), and divide by $2$:

$$\frac{18\cdot6\cdot9\cdot24\cdot11!}{2}=465,589,555,200$$

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ORIGINAL POST

I get the correct answer like this:

We have $13$ entities:

• 6 USA
• 5 French
• 1 Chinese conglomerate of $3$
• 1 Japanese conglomerate of $4$

There are $13!$ ways to arrange these, but then there is the additional rule that we cannot have either $JC$ or $CJ$.

There are $12\cdot11!$ permutations with $JC$ in them, and the same with $CJ$ in them, so the total number of permutations is $13!-2\cdot12!=(13-2)\cdot12!=11\cdot12!=5,269,017,600$.

Next multiply by $3!\cdot4!=144$ for the inner positioning of the Chinese and Japanese, and then a rectangular table has one line of symmetry, so divide by $2$, which gives $379,369,267,200$.