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

It is currently 15 Oct 2019, 22:45

Close

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
Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

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

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Find Similar Topics 
Manager
Manager
avatar
Joined: 16 Feb 2011
Posts: 164
Schools: ABCD
In how many ways can 5 boys and 3 girls be seated on 8  [#permalink]

Show Tags

New post Updated on: 24 Sep 2012, 13:37
3
27
00:00
A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

60% (02:11) correct 40% (02:33) wrong based on 230 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, 13:02.
Last edited by voodoochild on 24 Sep 2012, 13:37, edited 1 time in total.
Most Helpful Expert Reply
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58347
In how many ways can 5 boys and 3 girls be seated on 8  [#permalink]

Show Tags

New post 24 Sep 2012, 13:23
5
11
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.
_________________
General Discussion
Manager
Manager
avatar
Joined: 16 Feb 2011
Posts: 164
Schools: ABCD
Re: In how many ways can 5 boys and 3 girls  [#permalink]

Show Tags

New post 24 Sep 2012, 13:37
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 \(20*6*120=14,000\).

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
Manager
Manager
avatar
Joined: 16 Feb 2011
Posts: 164
Schools: ABCD
Re: In how many ways can 5 boys and 3 girls be seated on 8  [#permalink]

Show Tags

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

Arrangemnets = 6C3 * 5! * 3! = B...thanks
Director
Director
User avatar
Joined: 22 Mar 2011
Posts: 590
WE: Science (Education)
Re: In how many ways can 5 boys and 3 girls  [#permalink]

Show Tags

New post 24 Sep 2012, 14:32
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 \(20*6*120=14,000\).

Answer: B.


Small typo: \(20*6*120=14,000\) should be 14,400.
_________________
PhD in Applied Mathematics
Love GMAT Quant questions and running.
Director
Director
User avatar
S
Joined: 17 Dec 2012
Posts: 626
Location: India
In how many ways can 5 boys and 3 girls be seated on 8  [#permalink]

Show Tags

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.
_________________
Srinivasan Vaidyaraman
Sravna Test Prep
http://www.sravnatestprep.com

Holistic and Systematic Approach
Current Student
avatar
B
Joined: 21 Sep 2015
Posts: 5
Location: Singapore
GMAT 1: 710 Q49 V39
GPA: 3.75
Reviews Badge
Re: In how many ways can 5 boys and 3 girls be seated on 8  [#permalink]

Show Tags

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,
Math Expert
avatar
V
Joined: 02 Aug 2009
Posts: 7960
Re: In how many ways can 5 boys and 3 girls be seated on 8  [#permalink]

Show Tags

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..

_________________
Intern
Intern
avatar
Joined: 28 Dec 2015
Posts: 37
Re: In how many ways can 5 boys and 3 girls be seated on 8  [#permalink]

Show Tags

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?
Intern
Intern
avatar
Joined: 13 Jul 2016
Posts: 36
GMAT 1: 770 Q50 V44
GMAT ToolKit User
Re: In how many ways can 5 boys and 3 girls be seated on 8  [#permalink]

Show Tags

New post 03 Sep 2016, 11:02
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 \(20*6*120=14,400\).

Answer: B.


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
User avatar
V
Status: Stepping into my 10 years long dream
Joined: 18 Jul 2015
Posts: 3582
Reviews Badge
Re: In how many ways can 5 boys and 3 girls be seated on 8  [#permalink]

Show Tags

New post 03 Sep 2016, 11:13
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 \(20*6*120=14,400\).

Answer: B.


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 LinkedIn abhimahna.
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
Blog: Subscribe to Question of the Day Blog
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
Intern
avatar
Joined: 13 Jul 2016
Posts: 36
GMAT 1: 770 Q50 V44
GMAT ToolKit User
Re: In how many ways can 5 boys and 3 girls be seated on 8  [#permalink]

Show Tags

New post 03 Sep 2016, 11:29
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
Manager
avatar
B
Joined: 28 Jun 2016
Posts: 215
Location: Canada
Concentration: Operations, Entrepreneurship
Re: In how many ways can 5 boys and 3 girls be seated on 8  [#permalink]

Show Tags

New post 04 Sep 2016, 20:22
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 \(20*6*120=14,400\).

Answer: B.


# 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
User avatar
V
Status: Stepping into my 10 years long dream
Joined: 18 Jul 2015
Posts: 3582
Reviews Badge
In how many ways can 5 boys and 3 girls be seated on 8  [#permalink]

Show Tags

New post 04 Sep 2016, 23:53
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 \(20*6*120=14,400\).

Answer: B.


# 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 LinkedIn abhimahna.
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
Blog: Subscribe to Question of the Day Blog
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.
Senior Manager
Senior Manager
User avatar
P
Joined: 03 Mar 2017
Posts: 367
Reviews Badge CAT Tests
In how many ways can 5 boys and 3 girls be seated on 8  [#permalink]

Show Tags

New post 08 Jun 2019, 06:23
VeritasKarishma AjiteshArun egmat Bunuel
chetan2u

Why cannot we use the below approach ??

Total - when 2 girls sit together

Total--> 8!


When Two girls sit together-->> 5B 1G (G1G2)--> 7!*3C2*2!

Answer = (8!)-( 7!*3C2*2!)

What is wrong with this approach.
_________________
--------------------------------------------------------------------------------------------------------------------------
All the Gods, All the Heavens, and All the Hells lie within you.
Veritas Prep GMAT Instructor
User avatar
V
Joined: 16 Oct 2010
Posts: 9701
Location: Pune, India
Re: In how many ways can 5 boys and 3 girls be seated on 8  [#permalink]

Show Tags

New post 09 Jun 2019, 23:54
warrior1991 wrote:
VeritasKarishma AjiteshArun egmat Bunuel
chetan2u

Why cannot we use the below approach ??

Total - when 2 girls sit together

Total--> 8!


When Two girls sit together-->> 5B 1G (G1G2)--> 7!*3C2*2!

Answer = (8!)-( 7!*3C2*2!)

What is wrong with this approach.


I am not sure what you have done here - there are 3 girls and no two should sit together.
This means that G1 and G2 should not be together, G2 and G3 should not be together and G1 and G3 should not be together.

So you make the boys sit in 5 chairs. Now the girls can put their chairs in 6 spots -
S B1 S B2 S B3 S B4 S B5 S

You arrange the boys in 5! ways. Select 3 spots for girls in 6C3 ways and arrange the girls in 3! ways.

Total = 5! * 6C3 * 3! = 120 * 20 * 6 = 14400
_________________
Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
Senior Manager
Senior Manager
User avatar
P
Joined: 03 Mar 2017
Posts: 367
Reviews Badge CAT Tests
In how many ways can 5 boys and 3 girls be seated on 8  [#permalink]

Show Tags

New post 12 Jun 2019, 20:12
warrior1991 wrote:
VeritasKarishma AjiteshArun egmat Bunuel
chetan2u

Why cannot we use the below approach ??

Total - when 2 girls sit together

Total--> 8!


When Two girls sit together-->> 5B G (GG)--> 7!*3C2*2!

Answer = (8!)-( 7!*3C2*2!)

What is wrong with this approach.



VeritasKarishma chetan2u generis AjiteshArun Bunuel

Cannot we solve it like this??

Approach :- (Total cases - Cases when 2 girls sit together)

Total Cases :- 8!

When Two girls sit together-->> 5B G(GG)--> 7!*3C2*2!

Answer = (8!)-( 7!*3C2*2!)

However, I get different answer.

Can you please explain where am I going wrong in this case.
_________________
--------------------------------------------------------------------------------------------------------------------------
All the Gods, All the Heavens, and All the Hells lie within you.
Math Expert
avatar
V
Joined: 02 Aug 2009
Posts: 7960
Re: In how many ways can 5 boys and 3 girls be seated on 8  [#permalink]

Show Tags

New post 12 Jun 2019, 20:37
warrior1991 wrote:
warrior1991 wrote:
VeritasKarishma AjiteshArun egmat Bunuel
chetan2u

Why cannot we use the below approach ??

Total - when 2 girls sit together

Total--> 8!


When Two girls sit together-->> 5B G (GG)--> 7!*3C2*2!

Answer = (8!)-( 7!*3C2*2!)

What is wrong with this approach.



VeritasKarishma chetan2u generis AjiteshArun Bunuel

Cannot we solve it like this??

Approach :- (Total cases - Cases when 2 girls sit together)

Total Cases :- 8!

When Two girls sit together-->> 5B G(GG)--> 7!*3C2*2!

Answer = (8!)-( 7!*3C2*2!)

However, I get different answer.

Can you please explain where am I going wrong in this case.


You will get your answer lesser than actual because there is repetition in what you are subtracting.
Say girls are a,b,c.. so you choose B,B,(a,b),c,B,B...
When you choose b,c, one combination will again be B,B,a,(b,c),B,B...
So both are same but are subtracting it twice
_________________
GMAT Club Legend
GMAT Club Legend
User avatar
D
Joined: 18 Aug 2017
Posts: 4999
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)
GMAT ToolKit User Premium Member
Re: In how many ways can 5 boys and 3 girls be seated on 8  [#permalink]

Show Tags

New post 13 Jun 2019, 21:19
[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

[spoiler=]HEre's what I did :

girls can be made to sit
_b_b_b_b_b_
total 6 places to choose for girls ; 6c3 in 3! ways
and boys 5! ways
5!*6c3*3! = 14400
IMO B
Veritas Prep GMAT Instructor
User avatar
V
Joined: 16 Oct 2010
Posts: 9701
Location: Pune, India
Re: In how many ways can 5 boys and 3 girls be seated on 8  [#permalink]

Show Tags

New post 13 Jun 2019, 23:11
warrior1991 wrote:
warrior1991 wrote:
VeritasKarishma AjiteshArun egmat Bunuel
chetan2u

Why cannot we use the below approach ??

Total - when 2 girls sit together

Total--> 8!


When Two girls sit together-->> 5B G (GG)--> 7!*3C2*2!

Answer = (8!)-( 7!*3C2*2!)

What is wrong with this approach.



VeritasKarishma chetan2u generis AjiteshArun Bunuel

Cannot we solve it like this??

Approach :- (Total cases - Cases when 2 girls sit together)

Total Cases :- 8!

When Two girls sit together-->> 5B G(GG)--> 7!*3C2*2!

Answer = (8!)-( 7!*3C2*2!)

However, I get different answer.

Can you please explain where am I going wrong in this case.


As chetan2u mentioned above, you are double counting the eliminations. Whenever you need to arrange people such that no 2 sit together, one needs to arrange the others and place these people in between the others.
_________________
Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
GMAT Club Bot
Re: In how many ways can 5 boys and 3 girls be seated on 8   [#permalink] 13 Jun 2019, 23:11
Display posts from previous: Sort by

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

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  





Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne