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. 

One way to look at this is that there are $4$ possible places to seat Benjamin as per first condition. Now if we cannot seat him next to Grace, that leaves only $2$ possible places for him out of $4$. As there are equal number of seating arrangements for Benjamin being in each of these $4$ places,
Number of seating arrangements meeting both conditions $ = \displaystyle \frac{2}{4} \cdot 4! \cdot 3! = 72$
