Question 1207620
.
At a meeting, 4 scientists, 3 mathematicians, and 2 journalists are to be seated 
around a circular table. How many different arrangements are possible if every mathematician 
must sit next to a journalist? (Two seatings are considered equivalent if one seating 
can be obtained from rotating the other.)
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<pre>
Draw a circle - it will represent the circular table.


There are 9 chairs around the table.

Let assume that the chairs are numbered from 1 to 9 sequentially clockwise around the table 
and let assume that the chair #1 is in position "North", or 12 o'clock.


We will place one of the two journalist at the chair #1. 


Then the other journalist can not occupy neither of the two neighbor chairs,
since otherwise will be no place for 3 mathematicians next to two journalists.

It means that the other journalist can occupy any one chair from #3 to #8 inclusive.


Thus, there are 9 - 3 = 6 possibilities for the other journalist's chair.


OK. So, there are 6 possibilities to place two journalists.    (1)


Next, assume that two journalists are just placed this way.
Then there are 4 chairs neighbor these two journalists chairs to place 3 mathematician there.

There are  {{{C[4]^3}}} = 4 ways to place 3 mathematicians at these 4 chairs.    (2)


So, now we have 2+3 = 5 chairs occupied and 9-5 = 4 chairs free for four scientists.


These scientists can be placed in 4! = 24 different ways in these 4 chairs.    (3)


Now calculate the product of options

    n = 6 (from (1)) * 4 (from (2)) * 24 (from (3)) = 6 * 4 * 24 = 576.


At this point, the problem is solved to the end, and the number of 
all different arrangements is 576.    <U>ANSWER</U>
</pre>

Solved.


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It is very nice combinatorics problem of an Olympiad level.


I never saw and never solved similar combinatorics problems before.


It is very rare case to see a new, a fresh and so beautiful combinatorics problem (!)



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I agree with Edwin, noticing the missed factor 2.


So, the correct answer is 2*576 = 1152.


Thank you, Edwin !