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In how many ways can 5 boys and 3 girls be seated on 8

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In how many ways can 5 boys and 3 girls be seated on 8 [#permalink]

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New post 24 Sep 2012, 13:02
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In how many ways can 5 boys and 3 girls be seated on 8 chairs so that no two girls are together?

A. 5760
B. 14400
C. 480
D. 56
E. 40320

[Reveal] Spoiler:
HEre's what I did :

5! * (4C3) *2* 3! = 120*6*4*2= 5760. Am I correct?
[Reveal] Spoiler: OA

Last edited by voodoochild on 24 Sep 2012, 13:37, edited 1 time in total.
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In how many ways can 5 boys and 3 girls be seated on 8 [#permalink]

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voodoochild wrote:
In how many ways can 5 boys and 3 girls be seated on 8 chairs so that no two girls are together?


A 5760
B 14400
C 480
D 56
E 40320

HEre's what I did :

5! * (4C3) *2* 3! = 120*6*4*2= 5760. Am I correct?


Consider the following arrangement:

*B*B*B*B*B*

Now, if girls occupy the places of 6 stars no girls will be together.

# of ways 3 girls can occupy the places of these 6 stars is \(C^3_6=20\);
# of ways 3 girls can be arranged on these places is \(3!=6\);
# of ways 5 boys can be arranged is \(5!=120\).

So total # of ways to arrange 3 Girls and 5 boys so that no girls are together is \(20*6*120=14,400\).

Answer: B.
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Re: In how many ways can 5 boys and 3 girls [#permalink]

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New post 24 Sep 2012, 13:37
Bunuel wrote:
voodoochild wrote:
In how many ways can 5 boys and 3 girls be seated on 8 chairs so that no two girls are together?


A 5760
B 14400
C 480
D 56
E 40320

HEre's what I did :

5! * (4C3) *2* 3! = 120*6*4*2= 5760. Am I correct?


Consider the following arrangement:

*B*B*B*B*B*

Now, if girls occupy the places of 6 stars no girls will be together.

# of ways 3 girls can occupy the places of these 6 stars is \(C^3_6=20\);
# of ways 3 girls can be arranged on these places is \(3!=6\);
# of ways 5 boys can be arranged is \(5!=120\).

So total # of ways to arrange 3 Girls and 5 boys so that no girls are together is \(20*6*120=14,000\).

Answer: B.


Bunuel,

Here's what I thought:

_ O _ O _ O _ O

3 G can occupy any of the 4 "_" positions in 4C3 ways.

Similar Girls could also occupy any of the 4 "O" positions in 4C3 ways.

Boys can be permuted in 5! ways. Girls - 3! Therefore arrangements = 4C3* 2 * 5! * 3! = 4*2*120*6= 5760. Can you please let me know what I am missing?

Thanks
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Re: In how many ways can 5 boys and 3 girls be seated on 8 [#permalink]

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New post 24 Sep 2012, 13:59
Please ignore. I got it. 4C3 should be 6C3.

Arrangemnets = 6C3 * 5! * 3! = B...thanks
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Re: In how many ways can 5 boys and 3 girls [#permalink]

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New post 24 Sep 2012, 14:32
Bunuel wrote:
voodoochild wrote:
In how many ways can 5 boys and 3 girls be seated on 8 chairs so that no two girls are together?


A 5760
B 14400
C 480
D 56
E 40320

HEre's what I did :

5! * (4C3) *2* 3! = 120*6*4*2= 5760. Am I correct?


Consider the following arrangement:

*B*B*B*B*B*

Now, if girls occupy the places of 6 stars no girls will be together.

# of ways 3 girls can occupy the places of these 6 stars is \(C^3_6=20\);
# of ways 3 girls can be arranged on these places is \(3!=6\);
# of ways 5 boys can be arranged is \(5!=120\).

So total # of ways to arrange 3 Girls and 5 boys so that no girls are together is \(20*6*120=14,000\).

Answer: B.


Small typo: \(20*6*120=14,000\) should be 14,400.
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Re: In how many ways can 5 boys and 3 girls be seated on 8 [#permalink]

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Re: In how many ways can 5 boys and 3 girls be seated on 8 [#permalink]

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New post 21 Feb 2016, 05:44
Hello from the GMAT Club BumpBot!

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Re: In how many ways can 5 boys and 3 girls be seated on 8 [#permalink]

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New post 21 Feb 2016, 06:25
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Bunuel wrote:
voodoochild wrote:
In how many ways can 5 boys and 3 girls be seated on 8 chairs so that no two girls are together?


A 5760
B 14400
C 480
D 56
E 40320

HEre's what I did :

5! * (4C3) *2* 3! = 120*6*4*2= 5760. Am I correct?


Consider the following arrangement:

*B*B*B*B*B*

Now, if girls occupy the places of 6 stars no girls will be together.

# of ways 3 girls can occupy the places of these 6 stars is \(C^3_6=20\);
# of ways 3 girls can be arranged on these places is \(3!=6\);
# of ways 5 boys can be arranged is \(5!=120\).

So total # of ways to arrange 3 Girls and 5 boys so that no girls are together is \(20*6*120=14,000\).

Answer: B.


Hi Bunuel,
typo error,
it should be 14,400 and not 14000
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Re: In how many ways can 5 boys and 3 girls be seated on 8 [#permalink]

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New post 21 Feb 2016, 09:17
chetan2u wrote:
Bunuel wrote:
voodoochild wrote:
In how many ways can 5 boys and 3 girls be seated on 8 chairs so that no two girls are together?


A 5760
B 14400
C 480
D 56
E 40320

HEre's what I did :

5! * (4C3) *2* 3! = 120*6*4*2= 5760. Am I correct?


Consider the following arrangement:

*B*B*B*B*B*

Now, if girls occupy the places of 6 stars no girls will be together.

# of ways 3 girls can occupy the places of these 6 stars is \(C^3_6=20\);
# of ways 3 girls can be arranged on these places is \(3!=6\);
# of ways 5 boys can be arranged is \(5!=120\).

So total # of ways to arrange 3 Girls and 5 boys so that no girls are together is \(20*6*120=14,000\).

Answer: B.


Hi Bunuel,
typo error,
it should be 14,400 and not 14000


Edited. Thank you.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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In how many ways can 5 boys and 3 girls be seated on 8 [#permalink]

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New post 18 Mar 2016, 02:25
Let us take opposite of the constraint.

2 girls sitting together: :

1 case is GGBGBBBB.
Total number of ways=3!*5!*5 with just shifting the rightmost girl.
Then the 2 leftmost girls can shift one position , and using the above reasoning, the total number of ways = 3!*5!*4 and so on till the rightmost girl has 1 position.

So total number of ways = 3!*5!(5+4+3+2+1)=120*90=10800

Similarly another case is:
GBGGBBBB.
Using the above reasoning, the total number of cases is: 3!*5!*(15) =10800

Let us take 3 girls sitting together

GGGBBBBB
There are 3! *5! Ways. The 3 leftmost girls can shift 6 positions. So there are a total of 3!*5!*6=4320 ways

So total is 2*10800 + 4320=25920

The total number of possibilities = 8! Ways =40,320
Answer is 40320-25920=14400
Hence B.
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Re: In how many ways can 5 boys and 3 girls be seated on 8 [#permalink]

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New post 20 Mar 2016, 07:50
[quote="Bunuel"][quote="chetan2u"][quote="Bunuel"][quote="voodoochild"]In how many ways can 5 boys and 3 girls be seated on 8 chairs so that no two girls are together?


A 5760
B 14400
C 480
D 56
E 40320



Hi Bunuel,

What is the error in this solutions : 8! - 6!3!

Total ways - girls together = not together

Regards,
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Re: In how many ways can 5 boys and 3 girls be seated on 8 [#permalink]

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New post 20 Mar 2016, 08:44
urshila wrote:
Bunuel wrote:
chetan2u wrote:
In how many ways can 5 boys and 3 girls be seated on 8 chairs so that no two girls are together?


A 5760
B 14400
C 480
D 56
E 40320



Hi Bunuel,

What is the error in this solutions : 8! - 6!3!

Total ways - girls together = not together

Regards,


Hi,
the Q asks to negate even when two are together..
What you have used is only when all three are together..
that is why your answer, 36000, is more than the actual, 14400..

Hope you have realized where you have gone wrong..

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Re: In how many ways can 5 boys and 3 girls be seated on 8 [#permalink]

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New post 23 Jul 2016, 22:44
hello chetan2u@

I want to know how can it be done the other way round,subtracting from the total number of cases:
Say we have 3 girls:G1 G2 and G3.
In the first case,we can take G1 and G2 as a single element,so it will be B1B2B3B4B5G3(G1G2)=7!*2!
Similarly we can take for G2 and G3,considering them as a single element:B1B2B3B4B5G1(G2G3)=7!*2!
And for G1 and G3,considering them as a single element,we again have 7!*2!
Total number of cases=8!
So,Cases where no two girls are together=8!-7!*2!*3=10080.

I am unable to understand,what is wrong in this approach?
Re: In how many ways can 5 boys and 3 girls be seated on 8   [#permalink] 23 Jul 2016, 22:44
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