SOLUTION: In how many ways can 3 boys and 3 girls be seated at a round table if each girl is to be between two boys?

Considering circular permutation, we can assume that the chairs are numbered consequently 
from 1 to 6 clockwise and the chair number "1" is in the fixed position "North".


Then, if a girl is seated at the chair "1", then the sequence from 1 to 6 is "G B G B G B".

Making permutations inside the group of girls and inside the group of boys separately,
we have 3!*3!= 6*6 = 36 such different sequences, where girls occupy odd positions/chairs, 
while the boys occupy even positions/chairs.



Next, if a boy is seated at the chair "1", then the sequence from 1 to 6 is "B G B G B G".

Making permutations inside the group of girls and inside the group of boys separately, 
we have 3!*3!= 6*6 = 36 such sequences, where boys occupy odd positions/chairs, 
while the girls occupy even positions/chairs.



Formally, these 36 + 36 = 72 arrangements are different, since we can distinct them;

but as circular permutations, 36 seating arrangements of one type are equivalent to 36 seating arrangements of the other type,

so, actually, there are only 36 different circular permutations.



Thus, the final answer depends on which arrangements you call different.


There are 36 different circular permutations and 72 different arrangements, 
if we consider arrangements starting from girl or boy at the chair "1" as different.

Both the other two solutions are incorrect, according to the normal way 
circular-table enumeration problems are interpreted.

Ikleyn's answer of 36 would be correct if the table were assumed fixed 
so that it could not be rotated, then it would be the same as this 
unrotatable way, which is the way Ikleyn assumed the problem to be: 



There would be 3!=6 ways to place the boys and 3! ways to place the girls.
That would be (3!)(3!) = (6)(6) = 36 ways.  Then Ikleyn's answer would 
have been correct, with this assumption, which is in line with reality.

However, the creators of circular table math problems always assume that the
round table and the chairs are resting on a huge turntable and the huge
turntable can be rotated in any of the positions below. So all of these 6
seating positions below are considered by the problem creators to be just 1
position, because the imaginary huge turntable can be rotated to form any one 
of them: 



So we must divide the 36 ways by 6 and find that the solution intended by the
creators of the problem is only 6 ways.

Even though I realize that circular tables in reality are NEVER placed on
huge turntables, the people who make round-table math problems always
assume it anyway.  So we must always assume it when given a round-table 
problem, even though nobody puts tables and chairs on huge turntables.

The 6 ways are as follows. I will let P,Q, and R represent the three boys
and X, Y, and Z represent the 3 girls.  Then the 6 seating arrangements are
these:

 

Any of the other 30 seating arrangements of Ikleyn's answer of 36, can be
obtained by rotating the imaginary huge turntable on one of these 6. 

But the correct answer assumed by the creators is 6.  (Although I agree with
Ikleyn that the circular-table math problem creators are really the ones in
error).

Edwin