# Seating arrangements of 5 boys and 4 girls in a row, with no two girls adjacent

In how many ways five boys and four girls can be seated in a row so that no two girls are together?

• @user92907 What did you try so far? – rschwieb Sep 3 '13 at 15:23
• I'm new to this site. We chat about questions and do not solve it. Am I right?/ – user92907 Sep 3 '13 at 15:25
• I tried placing them alternatively – user92907 Sep 3 '13 at 15:26
• And yes, I deducted the number of ways by which 2 of them sit together times 2 , the number of ways 3 and 4 of them could sit together (times 3! and 4!) respectively from 9! – user92907 Sep 3 '13 at 15:28
• Helping you solve it requires some effort on your part. Math.SE isn't just a do-homework-for-lazy-students site, you understand. By providing your work so far, you proved you weren't one of those. Remember to post it with your next question, and you will get better results faster :) – rschwieb Sep 3 '13 at 16:12

First of all, the boys can be in any of $5!$ orders, if we ignore girls. Now imagine placing the girls after the boys are already in place. There are six spots for girls: Before the first boy, between the first and second, etc. The fact that no two girls can be adjacent means that no two can go in the same spot. The number of ways to place the girls is therefore $\tbinom 64\cdot4!$. This gives us a total of $\tbinom 64\cdot5!\cdot4! = \frac{6!5!}{2!} = 43200$.

Do those steps all make sense?

If the girls sit first, then they can be in any of $4!$ orders. There are now $5$ places, relative to the girls, to place boys: before the first girl, between girls $1$ and $2$, etc.

Not all of those $5$ places have to be filled with boys, but the middle three certainly do, and any of the places can have more than one boy. In fact, the numbers of boys in each of the five places can be any of the following sequences:

$1,1,1,1,1\\ 0,1,1,1,2\\ 0,1,1,2,1\\ 0,1,2,1,1\\ 0,2,1,1,1\\ 1,1,1,2,0\\ 1,1,2,1,0\\ 1,2,1,1,0\\ 2,1,1,1,0\\ 0,1,2,2,0\\ 0,2,1,2,0\\ 0,2,2,1,0\\ 0,3,1,1,0\\ 0,1,3,1,0\\ 0,1,1,3,0$

Those are $15$ options, and whichever one we choose, we can put the boys into their places in any of $5!$ orders.

Thus: $4!\cdot 15\cdot 5!$, just as before.

• what if girls sit first...would it make a difference?I am getting struck at this idea – Hydrous Caperilla Apr 22 '18 at 12:35
• Of course the girls can sit first, but the calculation gets more complicated. I'll make an edit to explain that version... – G Tony Jacobs Apr 22 '18 at 15:11
• Thanks for this explanation.This opens up a lot to me – Hydrous Caperilla Apr 23 '18 at 13:22

We need to seat them in such a way that the girls are not adjacent. To do this, let us first place all the boys in their seats.

Number of ways to do this = 5!

At this stage, it looks like this:

_B_B_B_B_B_

Now, we need to distribute the girls in the blanks. However, since the girls cannot be adjacent, we can only have a maximum of one girl in each blank.

There are 6 blanks and 4 girls, hence the number of ways to distribute them them is $^6C_4 \times 4!$ (because the order is important, we multiply by 4!).

Hence the total answer is $^6C_4 \times 4! \times 5!$

it should be 43200 because _b1_b2_b3_b4_b5_, boys will arrange themselves in 5! ways and the places left will be occupied by the girls in 6p4 ways. hence , answer= 5!*4!*6c4= 43200

• What do you mean by 6p4 and 6c4? Why are those factors here? What do they represent? – TZakrevskiy Sep 21 '13 at 15:38

(_B1_B2_B3_B4_B5_ ) for this arrangement $5$ boys take their seats in $5!=120$ ways. $4$ girls can take the $6$ positions in $^6P_4$ ways. So the total no of ways the boys and girls can sit so that no $2$ girls are together are $5! *$$^6P_4 = 120*360=43200.$

_B_B_B_B_B_ For boys, there are 5! possibility. For girls, there are 6 seats for them. the first girl can sit any of the six seats, second girl can sit any of the other 5 seats and so on. so 6x5x4x3.