Find all School-related info fast with the new School-Specific MBA Forum

It is currently 01 Aug 2014, 20:38

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.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

How many ways 3 girls and 4 boys can be seated so that girls

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Intern
Intern
avatar
Joined: 06 Apr 2004
Posts: 42
Followers: 0

Kudos [?]: 0 [0], given: 0

How many ways 3 girls and 4 boys can be seated so that girls [#permalink] New post 16 May 2005, 11:18
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
How many ways 3 girls and 4 boys can be seated so that girls never sit side by side?

My approach to this question is first to find total no. of arrangements, which is 7!. Then, first, subtract no. of possibilities when all 3 girls sit together (5!*3!) and, second, no. of possibilities when 2 girls out of 3 sit together.

Can someone help me to find the last item, i.e. no. of 2 girls out of 3 sitting together?

Unfortunately, I don't know the OA, so we need to find it on our own.
Manager
Manager
User avatar
Joined: 22 Apr 2005
Posts: 129
Location: Los Angeles
Followers: 1

Kudos [?]: 5 [0], given: 0

Re: PS: Permutation [#permalink] New post 16 May 2005, 11:38
You don't need to consider situations where three girls sit together.
You answer this question.
How many combinations are there where two girls sit together.
Make two girls as one.
Selecting two girls out of 3, ordered gives you 3!/1! = 3!
Then order the rest in 6! ways.
You end up with 3! * 6! ways.

The answer is 7! - 3! * 6! = 6! (7 - 6) = 6!
Senior Manager
Senior Manager
avatar
Joined: 15 Mar 2005
Posts: 425
Location: Phoenix
Followers: 1

Kudos [?]: 9 [0], given: 0

GMAT Tests User
Re: PS: Permutation [#permalink] New post 16 May 2005, 14:44
I am getting an answer of 5!x4x3 = 1440.

_ B _ B _ B _ B _
4 Boys can be rearranged in their prescribed positions in 4! ways.
3 Girls can be placed in 5 spaces in 5x4x3 ways.
Therefore, total = 4! x 5 x 4 x 3 = 5! x 4 x 3 = 1440.

Tyr, could you explain your working in detail please?
_________________

Who says elephants can't dance?

Manager
Manager
User avatar
Joined: 22 Apr 2005
Posts: 129
Location: Los Angeles
Followers: 1

Kudos [?]: 5 [0], given: 0

Re: PS: Permutation [#permalink] New post 16 May 2005, 16:28
kapslock wrote:
I am getting an answer of 5!x4x3 = 1440.

_ B _ B _ B _ B _
4 Boys can be rearranged in their prescribed positions in 4! ways.
3 Girls can be placed in 5 spaces in 5x4x3 ways.
Therefore, total = 4! x 5 x 4 x 3 = 5! x 4 x 3 = 1440.

Tyr, could you explain your working in detail please?


You have two times more outcomes then I do.

I think I see where's my flaw.
Classic
Counting number of outcomes where two girls sit together I counted arrangements of three girls together twice.
Number of outcomes where three girls sit together as noted originally - 5! * 3! = 6!
So number of outcomes where two girls sit together is in fact -
6! * 3! - 6!
And outcomes that we want 7! - 6! * 3! + 6! = 2 * 6!
Note to self, remember rules of Venn diagrams
Director
Director
avatar
Joined: 18 Feb 2005
Posts: 674
Followers: 1

Kudos [?]: 2 [0], given: 0

GMAT Tests User
 [#permalink] New post 16 May 2005, 19:57
4 boys can be arranged in 4! ways....

there are 5 spaces for the three girls ==>> 5C3 ways

Also the girls can be arranged in 3! ways among themselves.

4!*3!*5C3
Intern
Intern
avatar
Joined: 06 Apr 2004
Posts: 42
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: PS: Permutation [#permalink] New post 17 May 2005, 10:40
kapslock wrote:
I am getting an answer of 5!x4x3 = 1440.

_ B _ B _ B _ B _
4 Boys can be rearranged in their prescribed positions in 4! ways.
3 Girls can be placed in 5 spaces in 5x4x3 ways.
Therefore, total = 4! x 5 x 4 x 3 = 5! x 4 x 3 = 1440.

Tyr, could you explain your working in detail please?


kapslock, I've researched this type of problem today and come across the method used exactly the same as you described. Many thanks, guys, for insightful approaches.
Re: PS: Permutation   [#permalink] 17 May 2005, 10:40
    Similar topics Author Replies Last post
Similar
Topics:
Experts publish their posts in the topic In how many ways can 5 boys and 3 girls be seated on 8 voodoochild 4 24 Sep 2012, 12:02
1 Experts publish their posts in the topic Find the ways in which 4 boys and 4 girls can be seated mushyyy 2 21 Jan 2012, 10:01
1 Experts publish their posts in the topic In how many arrangements can a teacher seat 3 girls and 4 lumone 7 28 Nov 2007, 11:08
1 Experts publish their posts in the topic The music class consists of 4 girls and 7 boys. How many way bmwhype2 6 16 Oct 2007, 10:50
In how many ways 5 boys and 6 girls can be seated on 12 ps_dahiya 5 22 Jan 2006, 00:44
Display posts from previous: Sort by

How many ways 3 girls and 4 boys can be seated so that girls

  Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.