$8$ children of $4$ boys and $4$ girls are needed to form $2$ circles of $4$ children each. Find the number of different ways that this can be done.

Question

My method would be to have $${8 \choose 4} \times 3! \times {4 \choose 4} \times 3! \times 2! $$

Well I multiply the $2!$ behind because I think circle 1 and circle 2 can be interchanged. However the answer is $$\dfrac{{8 \choose 4} \times 3! \times {4 \choose 4} \times 3!}{2!} $$

Do not really understand why they need to divide by $2$.

Answer 1

Based on the solution given, this question is asking you to treat both circles equally. Thus, if you put people $A,B,C,D$ in circle $1$ and put people $E,F,G,H$ in circle $2$, this is the same as putting people $A,B,C,D$ in circle $2$ and people $E,F,G,H$ in circle $1$.

But the formula $\binom{8}{4}\times 3!\times\binom{4}{4}\times 3!$ treats these two arrangements as being distinct. Hence you must divide by $2$.