Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 16 Feb 2011
Posts: 192
Schools: ABCD

In how many ways can 5 boys and 3 girls be seated on 8 [#permalink]
Show Tags
Updated on: 24 Sep 2012, 13:37
3
This post received KUDOS
13
This post was BOOKMARKED
Question Stats:
67% (01:48) correct 33% (01:52) wrong based on 192 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?
Official Answer and Stats are available only to registered users. Register/ Login.
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.



Math Expert
Joined: 02 Sep 2009
Posts: 45497

In how many ways can 5 boys and 3 girls be seated on 8 [#permalink]
Show Tags
24 Sep 2012, 13:23
4
This post received KUDOS
Expert's post
7
This post was BOOKMARKED



Manager
Joined: 16 Feb 2011
Posts: 192
Schools: ABCD

Re: In how many ways can 5 boys and 3 girls [#permalink]
Show Tags
24 Sep 2012, 13:37
1
This post received KUDOS
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
Joined: 16 Feb 2011
Posts: 192
Schools: ABCD

Re: In how many ways can 5 boys and 3 girls be seated on 8 [#permalink]
Show Tags
24 Sep 2012, 13:59
Please ignore. I got it. 4C3 should be 6C3.
Arrangemnets = 6C3 * 5! * 3! = B...thanks



Director
Joined: 22 Mar 2011
Posts: 605
WE: Science (Education)

Re: In how many ways can 5 boys and 3 girls [#permalink]
Show Tags
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.
_________________
PhD in Applied Mathematics Love GMAT Quant questions and running.



Director
Joined: 17 Dec 2012
Posts: 635
Location: India

In how many ways can 5 boys and 3 girls be seated on 8 [#permalink]
Show Tags
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 4032025920=14400 Hence B.
_________________
Srinivasan Vaidyaraman Sravna http://www.sravnatestprep.com/bestonlinegrepreparation.php
Improve Intuition and Your Score Systematic Approaches



Intern
Joined: 21 Sep 2015
Posts: 5
Location: India
GPA: 3.75

Re: In how many ways can 5 boys and 3 girls be seated on 8 [#permalink]
Show Tags
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
Joined: 02 Aug 2009
Posts: 5784

Re: In how many ways can 5 boys and 3 girls be seated on 8 [#permalink]
Show Tags
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..
_________________
Absolute modulus :http://gmatclub.com/forum/absolutemodulusabetterunderstanding210849.html#p1622372 Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
GMAT online Tutor



Intern
Joined: 28 Dec 2015
Posts: 41

Re: In how many ways can 5 boys and 3 girls be seated on 8 [#permalink]
Show Tags
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
Joined: 13 Jul 2016
Posts: 41

Re: In how many ways can 5 boys and 3 girls be seated on 8 [#permalink]
Show Tags
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
Status: Stepping into my 10 years long dream
Joined: 18 Jul 2015
Posts: 3475

Re: In how many ways can 5 boys and 3 girls be seated on 8 [#permalink]
Show Tags
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 GMAT Story: From V21 to V40 My MBA Journey: My 10 years long MBA Dream My Secret Hacks: Best way to use GMATClub Verbal Resources: All SC Resources at one place  All CR Resources at one place
NEW > Subscribe to Question of the Day Blog
Find a bug in the new email templates and get rewarded with 2 weeks of GMATClub Tests for free



Intern
Joined: 13 Jul 2016
Posts: 41

Re: In how many ways can 5 boys and 3 girls be seated on 8 [#permalink]
Show Tags
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
Joined: 28 Jun 2016
Posts: 207
Location: Canada
Concentration: Operations, Entrepreneurship

Re: In how many ways can 5 boys and 3 girls be seated on 8 [#permalink]
Show Tags
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
Status: Stepping into my 10 years long dream
Joined: 18 Jul 2015
Posts: 3475

In how many ways can 5 boys and 3 girls be seated on 8 [#permalink]
Show Tags
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 \(C^3_6=20 How did you get this step?? Sent from my iPhone using GMAT Club Forum mobile appWe have 6 places for the girls to sit and we have only 3 girls. So, Girls could select the 3 places as [m]C^3_6=20\);
_________________
My GMAT Story: From V21 to V40 My MBA Journey: My 10 years long MBA Dream My Secret Hacks: Best way to use GMATClub Verbal Resources: All SC Resources at one place  All CR Resources at one place
NEW > Subscribe to Question of the Day Blog
Find a bug in the new email templates and get rewarded with 2 weeks of GMATClub Tests for free



NonHuman User
Joined: 09 Sep 2013
Posts: 6867

Re: In how many ways can 5 boys and 3 girls be seated on 8 [#permalink]
Show Tags
25 Nov 2017, 03:23
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.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources




Re: In how many ways can 5 boys and 3 girls be seated on 8
[#permalink]
25 Nov 2017, 03:23






