# Seating arrangement of $5$ men and $5$ women stand in a row so that no $2$ men and no $2$ women are adjacent to each other?

How many ways in which $5$ men and $5$ women stand in a row so that no $2$ men and no $2$ women are adjacent to each other?

• Thanks Arnaldo ..i have one doubt kindly see below.. In how many ways five boys and four girls can be seated in a row so that no two girls are together? There are 6 blanks and 4 girls, hence the number of ways to distribute them them is 6C4×4!6C4×4! (because the order is important, we multiply by 4!). Hence the total answer is 6C4×4!×5! ..so can u explain why it is different from previous asked question ? – Jeetu Jan 25 '17 at 14:41
• @Jeetu The question regarding five boys and four girls is only concerned with not seating two girls together. The reason the solution works is because we have $5$ boys who we seat and we have six slots $_B_B_B_B_B_$ for our girls. So then $GBBGBGBGB$ is a valid arrangement. If we tried to model the question you pose in this fashion we would arrive at $_B_B_B_B_B_$ and when we choose $5$ slots for our girls we are counting incorrect arrangements such as $GBGBBGBGBG$ when we skip a slot between selections. – ClownInTheMoon Jan 25 '17 at 14:59

Call $M$ for a men and $W$ for a women, so you want:

$$MHMHMHMHMH \quad \text{or} \quad HMHMHMHMHM$$

For both of them we have $5!5!$ (once men and women are differents) so the total is $2(5!)^2$

For your first seat you have $10$ individuals to choose from. For your second seat you cannot choose the same type of individual you chose for your first seat so you have $5$ ways to choose an individual for your second seat, for your third seat you have $4$ ways to choose an individual since you already chose one of this type of individual for your first seat... and so on.

We arrive at:

$$10 \cdot 5 \cdot 4 \cdot 4 \cdot 3 \cdot 3 \cdot 2 \cdot 2 \cdot 1 \cdot 1$$

or alternatively

$$10 \cdot 5! \cdot 4!.$$

To expand on the OP's comment. When we are concerned with arrangements of $4$ girls and $5$ boys such that no two girls are together, we can enumerate the possibilities by choosing $4$ slots out of the six slots that exist between our boys and then count all permutations of our boys and girls. Thus the answer is ${6 \choose 4} \cdot 5! \cdot 4!$.

However, in the question posed here, this argument does not work since if we try to do $6 \choose 5$ and place our girls between our five boys we will count arrangements where boys are next to one another since we don't have enough girls to guarantee that every slot between two boys is filled. Thus we need to fix an ordering and it is clear that we can either seat a girl first, or a boy first. We can look at either selecting an arbitrary person for the first seat (such as my model above) or we can look at cases and combine the results (such as the other answer did).

If one really desired a model involving combinations, you could model as follows. Let us seat our girls in this order where an $x$ denotes an empty seat: $xGxGxGxGxG$, now since we have exactly five seats and exactly five boys we can seat our boys in $5 \choose 5$ ways. Then for this arrangement we have ${5 \choose 5} \cdot 5! \cdot 5!$ valid arrangements. Now let us seat our girls as $GxGxGxGxGx$. Again we are in the clear here since we have $10$ seats total and enough boys to ensure there is one between every pair of girls. Then for this arrangement we have ${5 \choose 5} \cdot 5! \cdot 5!$. Combining our two possibilities we arrive at

$${5 \choose 5} \cdot 5! \cdot 5! +{5 \choose 5} \cdot 5! \cdot 5!= 2\cdot5!\cdot5!.$$