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

Question

There are 6 girls and 6 boys. If they are to be seated in a row, what is the probability that boys and girls sit alternately?

A)  \frac{1}{12!}

B)  \frac{{12C6}}{12!}

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

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

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

OA after two days

RDhin · Accepted Answer

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

As no. of Girls and Boys are same for sitting alternatively, one can start with a boy or a girl. { In case of 7 boys & 6 girls there can be only one arrangement in which boy can sit first }

1. Case

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

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

2. Case

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

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

Combining two

6!*6!+ 6!*6!= 2*6!*6!

So Probability is 2*6!*6!/12! Option D

FightToSurvive · Answer

There are 6 girls and 6 boys. If they are to be seated in a row, what is the probability that boys and girls sit alternately?

Its a tough one for me. As I don't like much of probability. But I will still give a try.

Total arrangement ways of 6 boys and 6 girls is 12!.

Imagine there is 12 seats. Then we can sit 6 boys and 6 girls alternatively in the below fashion
BGBGBGBGBGBG
GBGBGBGBGBGB
6ways*6ways*5ways*5ways*4ways*4ways*3ways*3ways*2ways*2ways*1way*1way = which is 6!*6! *2

Required probability is 2*6!*6!/12!

Its D IMO.

Kurtosis · Answer

Total number of ways in which 6 boys and 6 girls can sit in a row =  \frac{12!}{6!*6!}

Number of ways in which 6 boys and 6 girls can sit alternately = 2 (BGBG..... or GBGB.....)

Probability =  \frac{2 * 6! * 6!}{12!}

Answer: D

pvikas · Answer

please add oa to the stem.

dutchp0wner · Answer

I don't get it.

I see how it should it be 2 instead of 7, but why does the formula of the bold not hold?
*B*B*B*B*B*B*

# of ways 6 girls can occupy the places of these 7 stars is 6C7; 
# of ways 6 girls can be arranged on these places is 6!;
# of ways 6 boys can be arranged is 6!.

7∗6!∗6!/12! = D

Im not allowed to post links but the title of the topic is: A row of seats in a movie hall contains 10 seats. 3 Girls

Shivamraz · Answer

GMATNinja 
please help with it

Nirmesh83 · Answer

However , i understood the solution posted above by rest of users that in seat 1 - 6 boys can be seated, seat 2 -5 boys .. So , 6! ways boys can be seated . Similarly girls can be seated in 6! . Boys and Girls can be arranged in 2! ways
Probability =2x6!x6!/ 12!

What i don't understand is :-
We can assume all Girls are identical and boys are same. So , All 6 boys can be seated in 1 ways and rest 6 girls can be seated in 7 positions by 7C6 ways . So total no. of possible ways meeting constraint = 1x7C6x2= 14 ways

max. possible ways = ?? 
Probability = 14 / ??

Please correct my basic understanding ,where i am getting wrong.  Bunuel   VeritasKarishma   chetan2u  Guys please help  bb

KarishmaB · Answer

Nirmesh83

The girls are not identical and neither are the boys. People are all distinct. You need to arrange them. You have 12 distinct seats. 
On seat 1, you can make a girl sit or boy sit. This will decide which seats get occupied by girls and which by boys. If a girl sits on seat 1, seats 3, 5, 7, 9 and 11 will be taken by girls and seats 2, 4, 6, 8, 10 and 12 will be taken by boys. Then you arrange the girls and the boys in their own 6 seats each.

I am not sure from where 7 comes in. This is what I can think of. Say you are arranging 6 boys in 6 distinct seats in 6! ways. Now there are 7 open spots around them.

__ B __ B __ B __ B __ B __ B __

Note that we are not allowed to pick any 6 of these 7 spots so 7C6 is incorrect. If we do 7C6, acceptable arrangements will include:

G B G B G B __ B G B G B G
G B __ B G B G B G B G B G
etc..

But this will mean 2 boys are sitting together. That is not allowed.

Whether we pick the first slot or not fixes all other slots for us. If we pick the first slot for a girl, we get:
G B G B G B G B G B G B

If we do not pick the first slot for a girl, we get
B G B G B G B G B G B G

These are the only two acceptable options.

Nirmesh83 · Answer

VeritasKarishma  As per question , 2 boys should not sit together. People are distinct but we can classify as homogenous group on the basis of gender.
Thanks for pointing out the fact that _ will come between boys or girls which will void the constraint given in system.

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

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

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