There are 6 girls and 6 boys. If they are to be seated in a row, what

Question

Another Variant 
There are 6 girls and 6 boys. If they are to be seated in a row, what is the probability that no two girls sit together?

A)  \frac{1}{12!}

B)  \frac{{6!5!}}{12!}

C)  \frac{{6!6!}}{12!}

D)  \frac{2*{6!6!}}{12!}

E)  \frac{{6!7!}}{12!}

OA after two days

chetan2u · Accepted Answer

Hi..

Many of us have gone wrong on visualizing

If the Q asks you, no two boys or two girls sit together...
Means they sit alternate and therefore answer becomes \frac{2*6!*6!}{12!}

But here the restrictions are on two girls sitting together..
So place the boys in 6 alternate
 chairs..
Now there are 7 chairs one in beginning and other in end plus 5 in the gaps.
Choose 6 out of 7 =7C6=7
Arrange these in 6! Ways..
Ans 7*6!*6!/12!=7!*6!/12!
E

RDhin · Answer

Total no. of arrangement can be 12! as we have total 12 people in a sitting arrangement

In this case we will fix the positions of boys and tend to fill the girls in those vacant positions, because even if two boys are together we can still having an arrangement with two Girls sitting around the corners

So girls can occupy can spot in the following arrangement

Girls  .  Boys .  Girls  .  Boys .  Girls  .  Boys .  Girls  .  Boys .  Girls  .  Boys .  Girls  .  Boys .  Girls  .

total no. of arrangement in which girls can be filled is 7P6= 7!

total no. of arrangement in which boys can be filled is 6!

So total no. of arrangement in this case is 7!*6!

So Probability is 7!*6!/12! Option E

MasterPdog · Answer

But this arrangement has a total of 7 seats occupied by Girls. There are only 6 Girls.

I agree that  12!  is the total number of arrangements, so this goes in the denominator.

We also must agree that alternating seating Boy-Girl-Boy-Girl... (and Girl-Boy-Girl-Boy...) is the only way to avoid two girls seating next to each other. These are the 2 defined cases we need to permute: Boys sitting in odd seats, and boys sitting in even seats.

For case 1, up to 6 boys can take Seat 1. Any of 6 girls can take Seat 2. The number of available people in the second seat lowers by 1 to the end. So you have 6! boys times 6! girls.
Then you have to double the number for case 2, where a girl takes Seat 1 and boys take seat 2.

Therefore, option D seems correct.

RDhin · Answer

That analogy of 6 boys and 6 girls starting with any boy or girl makes senses when we are asked when boys and girls sit alternatively.

However in this case question we are explicitly asked for cases when no two girls sit together, however two boys can sit together.

So arrangement like this

G. B. B. G. B. G. B. G. B. G. B. G

G = Girl and B = Boy

Is valid, it means net arrangements will be more then case discussed by you as 2 boys can even be sit together.

That seven position significance is that one position among them will be empty. And that makes two boys sit together as it is exemplified above.

Sent from my Mi 4i using Tapatalk

rhine29388 · Answer

Can you please post the OA for the given question

aritragster · Answer

Can we have the OA for this question please?

gagan0001 · Answer

If girls are alternate, then only no 2 girls can be together.
So among 12 places, girls can be seated in 6 alternate places in 6! ways.
And boys can be seated in the remaining 6 places in 6! ways.
Therefore total number of ways = 6! * 6!
But there could be 2 such arrangements-->
1. When first place is filled by a girl
2. When first place is filled by a boy
Therefore total number of ways = 2*6! * 6!
Also total number of ways to fill 12 places = 12!
Hence probability = \frac{{2*6!6!}}{12!}

m1033512 · Answer

total combinations = 12 !

6 boys can sit = 6 ! ways
among them , there are 7 empty places where girls can 7P6 = 7 ! ways.

so probab = 6!*7!/12!

IMO E

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Mohammad Ali Khan · Answer

Chetan I am unable to pick its concept. pleasee elab. to help me understand. also if its given two girls do not sit together does that.. two or more girls arent sitting together i.e 3/4/5 and 6??

Fdambro294 · Answer

1st) DEN = Total arrangements of 12 people with no constraints in a row = 12!

2nd) NUM = No. of favorable arrangements in which no 2 girls are next to each other

line up the 6 boys with gaps around each

— B1 — B2 — B3 — B4 — B5 — B6 —

Because there must always be at least one Boy in between any 2 Girls in the Row, we need to choose 6 places out of 7 places in and around the 6 Boys

The number of ways we can do this is “7 choose 6” = 7 ways

3rd) then for each of the 7 ways, we can arrange the

-6 boys among themselves in 6! Ways

and

-the 6 girls among themselves in 6! Ways

7 * 6! * 6! = 7! * 6!

Prob = (7!) (6!) / (12!)

E

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There are 6 girls and 6 boys. If they are to be seated in a row, what

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There are 6 girls and 6 boys. If they are to be seated in a row, what

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