Question
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> 
> 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?