Number of ways they can be seated?

Question

3 boys $B_i ,\ i = 1, 2, 3$ and 6 girls $G_i, \ i = 1, 2, \ldots, 6$ are to be seated in a row. Number of ways they can be seated so that $B_1$ and $B_2$ are separated and $G_1$ and $G_2$ are also separated equal to?

I used the following approach:

Total no of ways - (No of ways $G_1$ , $G_2$ and $B_1$ , $B_2$ are together + No of ways $B_1$ , $B_2$ are together + No of ways $G_1$ , $G_2$ are together)

Which equals: 9! - (7!*2 *2 + 8!*2 + 8!*2)

Answer 1

Let $M$ be the event that number of ways $A_1$ and $A_2$ are together and $N$ be the event that number of ways $B_1$ and $B_2$ are together. Also, let $n(K)$ be the number of ways event $K$ can occur. Apply inclusion-exclusion principle to get, \begin{align*} \text{Required answer}&=9!-n(M\cup N)\\ &=9!-n(M)-n(N)+n(M\cap N)\\ &=9!-(2\times8!)-(2\times8!)+(4\times 7!)&&(\text{Using tie method})\\ &=7!(72-16-16+4)\\ &=\boxed{44\times7!} \end{align*}

Answer 2

You are close, but the number of ways G1 and G2 are together includes the cases where both G1 and G2 are together and B1 and B2 are together. Similarly the number of ways B1 and B2 are together includes the cases where both G1 and G2 are together and B1 and B2 are together. You have subtracted the case where both pairs are together twice, so you have to add it back in once. This is an example of the inclusion-exclusion principle.