# 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 $72$.

• how many seats? – Saketh Malyala Jun 14 '17 at 10:23
• Is this what you mean: "In how many ways can three girls and three boys sit in a row, if no two people of the same sex can be seated next to one another?" – user49640 Jun 14 '17 at 10:25
• Hundred of answers in just a minute – Brethlosze Jun 14 '17 at 10:31

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

• i know this approach .i want to know that what is wrong in my method! – laura Jun 14 '17 at 10:49
• @laura Well. I do not understand your $\binom{4}{3}$. Why $4$? The seats are $6$. – Robert Z Jun 14 '17 at 10:53
• because i have $4$ avaliable seats – laura Jun 14 '17 at 10:55
• @laura Why $4$ available seats? I am sorry but I do not understand this initial step. – Robert Z Jun 14 '17 at 10:57
• @laura Without the restriction for both boys and girls we have $\binom{6}{3}3!3!=6!=720$ ways. If the restriction holds only for boys then your approach is correct and the number of ways is 144. – Robert Z Jun 14 '17 at 11:40

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.

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

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. $$3!3!2=72$$