In how many ways 5 boys and 6 girls can be seated on 12

I will start solving this question from the Primary basic of Circular Arrangement (permutation).

In Circular Arrangement of Object we always fix one of the elements so that the repetition of arrangements (due to simultaneous movement of all objects in one of the directions) can be excluded

Here we Have 5 Boys, 6 Girls and 12 Chairs (Chairs Numbered from 1 to 12)

So Understand that one of the chairs will remain Vacant, Let's Fix the vacant chair only can call it CHAIR NO.1

Now, 6 Girls can sit only on chair no.s 2, 4, 6, 8, 10 and 12 only in 6! ways

And, 5 Girls can sit only on chair no.s 3, 5, 7, 9 and 11 only in 5! ways

i.e. Total ways of arranging all 11 people on 12 chairs with one chair vacant such that no boys sit together and no girls sit together = 5! * 6!

There are total 11 people and 12 chairs. Assume that V sits on the vacant chair. Now we have 12 chairs around a round table and 12 distinct "people".

Let's make the girls sit first.
One girl sits on any chair in 1 way (chairs around a table are not distinct relative to each other).
Now there are 11 distinct chairs (first to the girl's left, second to the girl's left, first to the girl's right etc).
Only 5 are available for the 5 girls - the chairs on either side of the girl are not available for girls. The girls can sit on only the alternate chairs. So 5 girls can sit on 5 distinct chairs in 5! ways.

Now 6 distinct chairs are leftover and 6 distinct people have to occupy them. This can be done in 6! ways.

Total number of arrangements = 1*5!*6! = 5! * 6!

Here are some posts on circular arrangements:

https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/10 ... angements/
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/10 ... ts-part-i/
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/11 ... nstraints-–-part-ii/

I think the questions says that
1) the arrangements of boys and girls
2) the seats on which they sit are different and have to be counted too( Although it actually doesn't make much difference, because the table is round)

x=boys
_=girls
0=free space

_x_x_x_x_x_0


We see that the free space cannot be betwen a girl and a boy, because otherwise a girl would sit next to another girl.

Fix the free seat, then

1) there are 6!*5!=86400 arrangements

adjust for 2) 6!*5!*12=1036800

Can't imagine that there are so many arrangements, will see what the others get.

Looks like new GMAT format with Quant section of 3 hours

6G & 5B

1) No. of ways 6 girls can sit on 12 chairs = 12P6
2) No. of ways in which any 2 girls sit together = 12P5

There are 6 chairs left,
3) No. of ways 5 boys can sit on 6 chairs = 6P5

So, total = (12P6 - 12P5) * 6P5

I come up with 6*5!*5!

No. of ways the girls can be seated on the round table with one vacant seat betwn all is (6-1)! = 5!
Now we have 6 vacant seats betn the girls-
select 5 seats out of 6 and arrange the boys=
5C6* 5!

so the no. of ways = 5!5!*6

I agree with allabout.

12*5!*6!

one empty seat - 12 ways of setting it up.
5 seats for 5 boys. - 5!
6 seats for 6 girls - 6!

hi,
most of us have gone wrong in the solution by multiplying the answer by 12..
as pointed by Karishma, the answer should be 6!5!..
reasons:
the 12 seats can be equally divided in 6 seats each ..
here 6 seats(alternate) are occupied by 6 boys. so these can be placed in 6! but since it is a circular table, the ways are (6-1)!=5!
and remaining 6 can be arranged in following ways... choosing 5 out of 6 =6 ways and arranging these 5 seats in 5! ways.. so total =6*5!=6!
total ways 6!5!..

the question is same as arranging 6 boys and 6 girls in 12 seats across a circular table... only that the vacant seat can be taken as a girl's seat..
However the solution changes say if we have two vacant seats or the number of boys is not half of total seats..
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It is a circular arrangement, so the order will not change if it is shifted.
We have a condition that no boy can sit by a girl, so the only arrangement can be

G B G B G B G B G B G _

B=5, so there are 5! ways to seat the boys
G=6, so there are 6! ways to seat the girls

Therefore there are 5!6! seating possibilities.

Shouldn't the answer be calculated as 5!*6C5*5!

[ girls can sit in (6-1)! ways creating 6 spaces in which 5 boys have to sit so 6C5 and finally 5! ways to arrange those boys]

I know the answer is the same but is the approach right?

Yes your approach is absolutely correct given that fact that you have considered that girls will sit on 6 alternate chairs in (6-1)! ways in order to leave space of exactly 1 chair between any two adjacent Girls.

Why does the solution become confusing all of a sudden if I place the boys first? It leaves 6 places for 6 girls and if we fix boys, as we placed them firstly, it gives (5-1)! = 4! for boys. Can you please help me get clear on this?

TheMastermind
May be i could help..


You see....the condition given in the question stem says that "no boy is seated adjacent to other boy and no girl is seated adjacent to other girl"...
So if you place the boys first then you have got only 5 places in a fixed circular table to accommodate 6 girls (You can not use the 12th chair as it has to remain vacant )...thus forcing at least 2 girls to sit together...and violating the condition in the stem.....so...in circular combinations.. it becomes a thumb rule when such a condition is given in the stem ... arrange the type with higher number first and then arrange other types around them ....


for arranging 6 girls on a fixed circular table...total no. of ways =(n-1)!=5!
Now ...you have got 6 places on a fixed circular table to accommodate 5 Boys.
So.. accommodating 5 boys in 6 available places...total no. of ways =6*5*4*3*2=6!
So.....total no. of ways=5!*6!

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