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!
Best Answer
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 $$2\cdot 3!\cdot 3!=72.$$