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

Question

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?

Answer 1

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 $6 \times 6 \times 2 = 72$