GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 09 Dec 2018, 21:16

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

## Events & Promotions

###### Events & Promotions in December
PrevNext
SuMoTuWeThFrSa
2526272829301
2345678
9101112131415
16171819202122
23242526272829
303112345
Open Detailed Calendar
• ### Free GMAT Algebra Webinar

December 09, 2018

December 09, 2018

07:00 AM PST

09:00 AM PST

Attend this Free Algebra Webinar and learn how to master Inequalities and Absolute Value problems on GMAT.
• ### Free lesson on number properties

December 10, 2018

December 10, 2018

10:00 PM PST

11:00 PM PST

Practice the one most important Quant section - Integer properties, and rapidly improve your skills.

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

Author Message
TAGS:

### Hide Tags

Manager
Joined: 16 Feb 2011
Posts: 180
Schools: ABCD
In how many ways can 5 boys and 3 girls be seated on 8  [#permalink]

### Show Tags

Updated on: 24 Sep 2012, 12:37
3
14
00:00

Difficulty:

55% (hard)

Question Stats:

66% (02:07) correct 34% (02:38) wrong based on 215 sessions

### HideShow timer Statistics

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?

Originally posted by voodoochild on 24 Sep 2012, 12:02.
Last edited by voodoochild on 24 Sep 2012, 12:37, edited 1 time in total.
Math Expert
Joined: 02 Sep 2009
Posts: 51035
In how many ways can 5 boys and 3 girls be seated on 8  [#permalink]

### Show Tags

24 Sep 2012, 12:23
4
9
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 $$20*6*120=14,400$$.

_________________
##### General Discussion
Manager
Joined: 16 Feb 2011
Posts: 180
Schools: ABCD
Re: In how many ways can 5 boys and 3 girls  [#permalink]

### Show Tags

24 Sep 2012, 12:37
1
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 $$20*6*120=14,000$$.

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
Manager
Joined: 16 Feb 2011
Posts: 180
Schools: ABCD
Re: In how many ways can 5 boys and 3 girls be seated on 8  [#permalink]

### Show Tags

24 Sep 2012, 12:59
Please ignore. I got it. 4C3 should be 6C3.

Arrangemnets = 6C3 * 5! * 3! = B...thanks
Director
Joined: 22 Mar 2011
Posts: 601
WE: Science (Education)
Re: In how many ways can 5 boys and 3 girls  [#permalink]

### Show Tags

24 Sep 2012, 13:32
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 $$20*6*120=14,000$$.

Small typo: $$20*6*120=14,000$$ should be 14,400.
_________________

PhD in Applied Mathematics
Love GMAT Quant questions and running.

Director
Joined: 17 Dec 2012
Posts: 630
Location: India
In how many ways can 5 boys and 3 girls be seated on 8  [#permalink]

### Show Tags

18 Mar 2016, 01:25
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
Hence B.
_________________

Srinivasan Vaidyaraman
Sravna Holistic Solutions
http://www.sravnatestprep.com

Holistic and Systematic Approach

Intern
Joined: 21 Sep 2015
Posts: 5
Location: Singapore
GMAT 1: 710 Q49 V39
GPA: 3.75
Re: In how many ways can 5 boys and 3 girls be seated on 8  [#permalink]

### Show Tags

20 Mar 2016, 06:50
[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,
Math Expert
Joined: 02 Aug 2009
Posts: 7095
Re: In how many ways can 5 boys and 3 girls be seated on 8  [#permalink]

### Show Tags

20 Mar 2016, 07:44
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..

_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

GMAT online Tutor

Intern
Joined: 28 Dec 2015
Posts: 39
Re: In how many ways can 5 boys and 3 girls be seated on 8  [#permalink]

### Show Tags

23 Jul 2016, 21:44
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?
Intern
Joined: 13 Jul 2016
Posts: 37
GMAT 1: 770 Q50 V44
Re: In how many ways can 5 boys and 3 girls be seated on 8  [#permalink]

### Show Tags

03 Sep 2016, 10:02
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 $$20*6*120=14,400$$.

I am not sure in this kind of problems why don't we consider the following arrangements : BGBBGBGB, BBGBGBGB etc when two of the boys are together
Board of Directors
Status: Stepping into my 10 years long dream
Joined: 18 Jul 2015
Posts: 3614
Re: In how many ways can 5 boys and 3 girls be seated on 8  [#permalink]

### Show Tags

03 Sep 2016, 10:13
siddharthharsh 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 $$20*6*120=14,400$$.

I am not sure in this kind of problems why don't we consider the following arrangements : BGBBGBGB, BBGBGBGB etc when two of the boys are together

Notice that we have already included these scenarios out of 14,400.

See, Bunuel has already included 5 places for Boys and 6 places for Girls, while we have 8 people in total.
_________________

My GMAT Story: From V21 to V40
My MBA Journey: My 10 years long MBA Dream
My Secret Hacks: Best way to use GMATClub | Importance of an Error Log!
Verbal Resources: All SC Resources at one place | All CR Resources at one place

GMAT Club Inbuilt Error Log Functionality - View More.
New Visa Forum - Ask all your Visa Related Questions - here.
New! Best Reply Functionality on GMAT Club!
Find a bug in the new email templates and get rewarded with 2 weeks of GMATClub Tests for free
Check our new About Us Page here.

Intern
Joined: 13 Jul 2016
Posts: 37
GMAT 1: 770 Q50 V44
Re: In how many ways can 5 boys and 3 girls be seated on 8  [#permalink]

### Show Tags

03 Sep 2016, 10:29
abhimahna wrote:

Notice that we have already included these scenarios out of 14,400.

See, Bunuel has already included 5 places for Boys and 6 places for Girls, while we have 8 people in total.

Thanks. So the logic is whenever the space that was left out for girls to occupy is vacant then the two adjacent boys are actually together. Was kind of difficult to see unless I formulated it in my own words. That is the trick with P&C and probability, kind of seems obvious if you have nailed it, but requires a bit of imagination in some easy ones too.
Manager
Joined: 28 Jun 2016
Posts: 207
Concentration: Operations, Entrepreneurship
Re: In how many ways can 5 boys and 3 girls be seated on 8  [#permalink]

### Show Tags

04 Sep 2016, 19:22
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 $$20*6*120=14,400$$.

# of ways 3 girls can occupy the places of these 6 stars is [m]C^3_6=20

How did you get this step??

Sent from my iPhone using GMAT Club Forum mobile app
Board of Directors
Status: Stepping into my 10 years long dream
Joined: 18 Jul 2015
Posts: 3614
In how many ways can 5 boys and 3 girls be seated on 8  [#permalink]

### Show Tags

04 Sep 2016, 22:53
acegmat123 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 $$20*6*120=14,400$$.

# of ways 3 girls can occupy the places of these 6 stars is $$C^3_6=20 How did you get this step?? Sent from my iPhone using GMAT Club Forum mobile app We have 6 places for the girls to sit and we have only 3 girls. So, Girls could select the 3 places as [m]C^3_6=20$$;
_________________

My GMAT Story: From V21 to V40
My MBA Journey: My 10 years long MBA Dream
My Secret Hacks: Best way to use GMATClub | Importance of an Error Log!
Verbal Resources: All SC Resources at one place | All CR Resources at one place

GMAT Club Inbuilt Error Log Functionality - View More.
New Visa Forum - Ask all your Visa Related Questions - here.
New! Best Reply Functionality on GMAT Club!
Find a bug in the new email templates and get rewarded with 2 weeks of GMATClub Tests for free
Check our new About Us Page here.

Non-Human User
Joined: 09 Sep 2013
Posts: 9084
Re: In how many ways can 5 boys and 3 girls be seated on 8  [#permalink]

### Show Tags

25 Nov 2017, 02:23
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: In how many ways can 5 boys and 3 girls be seated on 8 &nbs [#permalink] 25 Nov 2017, 02:23
Display posts from previous: Sort by