Question 1194841
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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?
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<pre>
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.
</pre>

Solved.



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I agree with the analysis by Edwin.


So, use his solution - it is correct,

and ignore my solution as wrong.