How many ways can $3$ boys and $3$ girls sit in a row if the boys and girls are each to sit together? Question

How many ways can $3$ boys and $3$  girls sit in a row if the boys and the girls are each to sit together?

My Approach

Total number of students =$6$,
  If we consider each student as one cell then total arrangements is $6!$.  Now it is given that girls and boys must sit together.
   So after taking $G$$B$ or $B$$G$ together we are left with 3 cell .
Total number of ways =$3!*2^3=48$,

$2$ option for each pair i.e either $GB$ or $BG$ and we have $3$ pair but the answer is given $72$
Please help me out.
Thanks!
Edit:  The full question from A First Course in Probability by Sheldon Ross reads:
(a) In how many ways can $3$ boys and $3$ girls sit in a row?
(b) In how many ways can $3$ boys and $3$ girls sit in a row if the boys and girls are each to sit together?
(c) In how many ways if only the boys sit together?
(d) In how many ways if no two people of the same sex sit together?
 A: To get $72$ interpreting this as the boys sitting together and the girls sit together:


*

*$3!=6$ ways of ordering the boys among themselves

*$3!=6$ ways of ordering the girls among themselves

*$2!=2$ ways of ordering the two groups


Then $6 \times 6 \times 2 = 72$
A: The question assumes that boys and girls alternate. As such, we can have $BGBGBG$ and $GBGBGB$. There are 3 options for the first boy and girl, 2 for the second and 1 for the last, so the number of possible arrangements equals:
$$2 \cdot 3! \cdot 3! = 72$$
Edit: the question is indeed not clear at all. According to the solution file provided by Toby Mak, they meant to ask in how many ways the boys and girls can sit in a row, with all girls sitting together and all boys sitting together. In this case we can have $BBBGGG$ or $GGGBBB$, which again results in $2 \cdot 3! \cdot 3! = 72$ possible arrangements.
A: Another way you can think about the problem with boys and girls alternating is with the n choose k (n,k) method.
My idea was to group a boy and a girl into one thing. Then we have three entities we want to sort into three seats. 
Then the answer becomes:
(ways of grouping group 1)(ways of seating group 1) + (ways of grouping group 2)(ways of seating group 2) + (ways of grouping group 3)(ways of seating group 3)
= 
(which gender first)(3 boys choose 1)(3 girls choose 1)(3 possible groups to sit in seat 1)
 +
(which gender first)(2 boys left, choose 1)(2 girls left choose 1)(2 possible groups to sit in seat 2) 
+ 
(which gender first)(1 boy left, choose 1)(1 girl left, choose 1)(1 possible group left to sit in seat 3)
(2,1)(3,1)(3,1)(3) + (2,1)(2,1)(2,1)(2) + (2,1)(1,1)(1,1)(1) = 72
