Lesson Upper level problems on a party of people sitting at a round table

Upper level problems on a party of people sitting at a round table

Problem 1

In such problems, the indistinguishable arrangements are those that are obtained 
one from the other by a circular rotation. 


For 10 people around a round table, there are 9! distinguishable arrangements 
(= a universal set U of arrangements).

Of them, there are 8! arrangements, where A sits next to C on the left of him
and 8! arrangements, where A sits next to C on the right of him.


So, in the universal set U, there are 2*8! arrangements {A}, where A sits next to C.


Similarly, in the universal set U, there are 2*8! arrangements {B}, where B sits next to C.


These arrangements, {A} and {B}, have non-empty intersection.


This intersection has 7! arrangements {ACB} and 7! other arrangements {BCA}.


From it, we conclude that the number of favorable arrangements is 

    9! - 2*8! - 2*8! + 2*7!  = 1*2*3*4*5*6*7*8*9 - 4*(1*2*3*4*5*6*7*8) + 2*(1*2*3*4*5*6*7) = 211680.    ANSWER

Problem 2

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 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 two neighbor chairs,
since otherwise there is 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 2*6 = 12 possibilities to place two journalists.    (1)


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

There are   = 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.


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


Now calculate the product of options

    n = 12 (from (1)) * 4 (from (2)) * 24 (from (3)) = 12 * 4 * 24 = 1152.


At this point, the problem is solved to the end, and the number of 
all different arrangements is 1152.    ANSWER