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. 
 A: "I subtracted $6$ because it was supposedly the number of possibilities of Benjamin and Grace sitting next to each other"
This number was too small.  $6$ is just the number of places that you could seat Benjamin and Grace in the row... it being B G b g b g b, or b G B g b g b or b g B G b g b or ... or b g b g b G B where the uppercase B stands for where Benjamin sits and G where Grace sits.  The lowercase b's and g's still need to be determined which person it was specifically who was sitting there before we have uniquely determined an overall possibility.
Once having chosen which of the six possibilities for how Ben and Grace sit, we then from left-to-right decide which other boys sit in each of the seats designated for boys and which other girls sit in the seats designated for girls.  This can be done in $3!$ and $2!$ ways respectively for a total of $6\cdot 3!\cdot 2!=72$ overall outcomes for how Ben and Grace can sit.
Our grand total is then
$$4!\cdot 3! - 6\cdot 3!\cdot 2! = 144-72=72$$
A: 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 $12 \times 3! =72$
A: 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$
