In how many ways if no two people of the same sex are allowed to sit together? Question

In how many ways if no two people of the same sex are allowed to sit together if there are $3$ boys and $3$ girls?

My Approach
I found out all possible seats of the girls after fixing boy's seat
$$-B-B-B-$$
After fixing boy's seat  I have $\binom{4}{3}$ ways to select possible seat for girls.
After seat has been allocated to both boys and girls
there are $3!$ ways for girls to change seat among themselves
and $3!$ ways for boys to change seat among themselves.
Total number of ways is $\binom{4}{3}\cdot 3!\cdot 3!=144$,
but the answer is given as $72$.
Where am I wrong? Please help me out! Thanks!
 A: I guess you are considering $6$ seats in a row. We can start with a boy or a girl and then sex is alternated i.e. $BGBGBG$ or $GBGBGB$ (2 ways). We have $3!$ ways to arrange the boys and $3!$ ways to arrange the girls. Hence the total number of ways is
$$2\cdot 3!\cdot 3!=72.$$
A: You can have Boy-Girl-Boy-Girl-Boy-Girl or Girl-Boy-Girl-Boy-Girl-Boy
You can arrange the boys in $3!$ ways and the girls in $3!$ ways.
The answer is thus $2\cdot 3!\cdot 3!=\boxed{72}$ ways.
A: The groups of different three girls into different seats:
$$
3!
$$
And the group of three different boys into different seats:
$$
3!
$$
And the way they can sit is just:
$$
2
$$
because they are interlaced $BGBGBG$ or $GBGBGB$. No other ways are possible when two $GG$ or $BB$ are allowed.
The result is 72 as requested.
$$
3!3!2=72
$$
A: You have six seats so the two possible combinations are:

Boy Girl Boy Girl Boy Girl
Girl Boy Girl Boy Girl Boy

For each arrangement, you have 3! ways to arrange the boys and 3! ways to arrange the girls. Therefore the result is:
Number of possible combinations * Number of ways to arrange the boys * Number of ways to arrange the girls

Value: $$2 *3!*3! = 72$$
