In how many ways can $4$ boys and $4$ girls be seated in $2$ concentric circles if there are four in each circle?

Question

Question: Four boys, four girls sit around 2 concentric circles such that there are four in each circle. In how many ways can they be seated if there are no restrictions?

My Attempt: $2\cdot(^8C_4 \cdot 3! \cdot 3!)$ First I chose $4$ people out of the $8$, then next the group of $4$ can choose either circle, hence the multiplication of $2$. Then in the inner circle, the 4 people can be arranged $3!$ and then in the outer circle, the 4 people can also be rearranged $3!$, hence the $3!3!$.

Offical Answer: $2\cdot7!$ The textbook explanation was inadequate to me, and simply said the first person can choose either 2 circles to go into.

My Problem: I do not understand why it is simply $7!$. I understand if there was just 1 circle of $8$, then the possibility would be $7!$. However with $2$ circles wouldn't you have to consider the fact that either circles have to be reduced.

Answer 1

1. The first person choose either 2 circles to go into.
2. Now, the 2 circles become different (one circle with person 1, and the other without person 1)
3. Therefore, there are 7 distinct seats remaining among the 2 circles, so, there are $7!$ number to assign the remaining 7 people to the seats.

In conclusion, the total number of ways $= 2*7!$

Answer 2

Your intuition is almost correct with a little fault.

When you arrange the people on one circle then the seats on other circle become all distinguishable (With respect to the other circle where you arrange first 4 people) and hence linear permutations needs to be considered instead of circular permutations. Hence using this intuition the answer comes out to be

$$\binom {8}{4} . 3!. 4! = 2 . 7! $$

Consider the the people to be $A, B, C, D, 1,2,3,4.$ Now consider the two cases below

Now in the inner circle if we consider the circular permutations the cyclic order of $(A, B, C, D)$ is same as $(B, C, D, A) $. But now when we focus on both circles together, even after keeping the arrangement of people on outer circle the same, both the cases differ relative to their positions with the people on inner circle.

This is why when you wrote down the cyclic permutations in your solution, you missed out some cases which should not have been rejected. Consider the example I gave. Your solution counts both these ways as same but in fact they are different.

Hope it helped.