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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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 \(20*6*120=14,400\).

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

Show Tags

24 Sep 2012, 12:37

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?

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 \(20*6*120=14,000\).

Answer: B.

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

PhD in Applied Mathematics Love GMAT Quant questions and running.

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

Show Tags

08 Jan 2015, 00:59

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 [#permalink]

Show Tags

21 Feb 2016, 04:44

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.
_________________

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
_________________

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

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 Answer is 40320-25920=14400 Hence B.
_________________

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?

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.. _________________

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?

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 \(20*6*120=14,400\).

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

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 \(20*6*120=14,400\).

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.
_________________

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.

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 \(20*6*120=14,400\).

Answer: B.

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

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 \(20*6*120=14,400\).

Answer: B.

# of ways 3 girls can occupy the places of these 6 stars is \(C^3_6=20

Happy New Year everyone! Before I get started on this post, and well, restarted on this blog in general, I wanted to mention something. For the past several months...

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Post-MBA I became very intrigued by how senior leaders navigated their career progression. It was also at this time that I realized I learned nothing about this during my...