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3 boys and 3 girls

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3 boys and 3 girls [#permalink] New post 19 Aug 2011, 22:59
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in how many different ways can 3 boys and 3 girls be seated in a row of 6 chairs such that the girls are not separated and boys are not separated ?

i have no idea how to solve this question. Please Explain

Thanks in advance
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Re: 3 boys and 3 girls [#permalink] New post 19 Aug 2011, 23:15
DeeptiM wrote:
Is 20 the answer?


i'm assuming that you tried 6C3 = 6!/3!*3! = 20. i don't know if its right or wrong because if this problem is thought in terms of slotting method i don't think, there could be 20 ways. because BBB GGG ... you can't separate them . and you have six chairs. Now i don't know if the chairs are placed next to each other or in circular shape. . . :roll:
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Re: 3 boys and 3 girls [#permalink] New post 20 Aug 2011, 14:34
girls and boys can be separated only through the following arrangements.

B B B G G G

G G G B B B


total arrangements = 6!/(3! 2!) = 20

total arrangements in which girls and boys are not separated = total arrangements - total arrangements in which they
are separated.

= 20 - 2 =18

What is the answer?
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Re: 3 boys and 3 girls [#permalink] New post 20 Aug 2011, 15:25
Since Boys and Girls have to be together, we have two consider them as two groups:

Boys can be arranged in 3! ways.
Girls can be arranged in 3! ways.
Group can be arranged in 2! ways.

So the total no of ways should be 3!*3!*2!=72.
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Re: 3 boys and 3 girls [#permalink] New post 22 Aug 2011, 20:34
Spidy001 wrote:
girls and boys can be separated only through the following arrangements.

B B B G G G

G G G B B B


total arrangements = 6!/(3! 2!) = 20

total arrangements in which girls and boys are not separated = total arrangements - total arrangements in which they
are separated.

= 20 - 2 =18

What is the answer?



isn't it 20 the correct answer. why 2 has been subtracted from 20?

i don't have OA
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Re: 3 boys and 3 girls [#permalink] New post 22 Aug 2011, 20:37
nish21in wrote:
Since Boys and Girls have to be together, we have two consider them as two groups:

Boys can be arranged in 3! ways.
Girls can be arranged in 3! ways.
Group can be arranged in 2! ways.

So the total no of ways should be 3!*3!*2!=72.



it can be 72 . i guess
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Last edited by shrive555 on 13 Oct 2011, 08:54, edited 1 time in total.
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Re: 3 boys and 3 girls [#permalink] New post 22 Aug 2011, 20:54
shrive555 wrote:
nish21in wrote:
Since Boys and Girls have to be together, we have two consider them as two groups:

Boys can be arranged in 3! ways.
Girls can be arranged in 3! ways.
Group can be arranged in 2! ways.

So the total no of ways should be 3!*3!*2!=72.



it can't be 72 . i guess


shrive555 is right..!!
Answer - 3!*3!*2!=72.

Cheers.!
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Re: 3 boys and 3 girls [#permalink] New post 22 Aug 2011, 20:56
I second 72..
total ways to arrange boys : 3!
total ways of arrange girls : 3!
if they are not to be seperated : 2!
total combinations : 3! * 3! *2! = 72

anyone who has not got the same ans, can u please explain what wrong here?
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Re: 3 boys and 3 girls [#permalink] New post 23 Aug 2011, 02:00
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Hello,
i would solve it this way.

1. All we have now is 1 group of boys and one group of girls : so total arrangement possible is 2! = 2
and
2. Group of boys (3)can be arranged in 3! ways : so total arrangements are 3! = 6
and
3. group of girls can be arranged in 3! ways : so total arrangements are 3! = 6

so the total number of ways are 2*6*6 = 72

This type of solving is also applicable for questions that ask for arranging a couple who cannot sit seperated.


Consider giving me kudos if my explanantion was helpful. :-D
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Re: 3 boys and 3 girls [#permalink] New post 14 Sep 2011, 12:13
Lets assume

b1,b2,b3 g1,g2,g3

so we have this case where they cant be separated so

we have only two combinations

1) b1b2b3g1g2g3 --Arranged in 3! *3! ways = 36
2) g1g2g3b1b2b3 --Arranged in 3! *3! ways = 36


Total 72 ways
Re: 3 boys and 3 girls   [#permalink] 14 Sep 2011, 12:13
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