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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 [#permalink]

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New post 29 Mar 2016, 04:36
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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
[Reveal] Spoiler: OA

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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

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Re: There are 6 girls and 6 boys. If they are to be seated in a row, what [#permalink]

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New post 29 Mar 2016, 05:16
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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

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Re: There are 6 girls and 6 boys. If they are to be seated in a row, what [#permalink]

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New post 29 Mar 2016, 10:44
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RDhin wrote:
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


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.

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Re: There are 6 girls and 6 boys. If they are to be seated in a row, what [#permalink]

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New post 29 Mar 2016, 12:36
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MasterPeeNice wrote:
RDhin wrote:
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


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.

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.

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Re: There are 6 girls and 6 boys. If they are to be seated in a row, what [#permalink]

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New post 03 Jul 2016, 14:26
chetan2u wrote:
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

Can you please post the OA for the given question

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Re: There are 6 girls and 6 boys. If they are to be seated in a row, what [#permalink]

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New post 03 Jul 2016, 20:00
Can we have the OA for this question please?

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Re: There are 6 girls and 6 boys. If they are to be seated in a row, what [#permalink]

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New post 03 Jul 2016, 22:27
chetan2u wrote:
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



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 =\(\frac{{2*6!6!}}{12!}\)

Kudos [?]: 20 [0], given: 8

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Re: There are 6 girls and 6 boys. If they are to be seated in a row, what [#permalink]

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New post 01 Dec 2017, 08:25
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chetan2u wrote:
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


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
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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

Kudos [?]: 6140 [1], given: 121

Re: There are 6 girls and 6 boys. If they are to be seated in a row, what   [#permalink] 01 Dec 2017, 08:25
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