How many possible seat arrangements are there? 
Four boys Bobby, Benson Benny, Benjamin and three girls Grace, Gloria, Georgia are to be seated in a row according to the following rules:

*

*a boy will not sit next to another boy and a girl will not sit next to another girl.

*Benjamin must not sit next to Grace. How many possible seat arrangements are there?


My solution:
$$3!4! - 6 = 138$$
The $3!4!$ is the answer if the rule is only #$1$. I subtracted $6$ because it’s supposedly the number of possibilities Benjamin and Grace sitting next to each other.
I’m sure there are a lot of flaws in my solutions since in the answer key, the answer is $72$.
I have just recently dived into these type of problems so please give me constructive criticism for the many flaws and preferably some detailed explanations. Thank you. 

Let name them $B_1,B_2,B_3,B_4,G_1,G_2,G_3$ ,respectively.
Let arrange girls such that $-G_1-G_2-G_3-$ , there are $4$ place to place boys and we do not want $G_1,B_4$ together.
So , there are $3$ possible choices for left of $G_1$ and $2$ choices for the right of it (Because we cannot place $B_4$ right or left of $G_1$ so we have $3 \times 2$ boys to place next to her).
Moreover , when you place boys like that , there are  $3 \times 2 \times 2 \times1 = 12$ arrangement for boys and there are $3!$ for girls .
So $12 \times 3! =72$
