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

It is currently 16 Apr 2014, 02:19

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

six students are equally divided into 3 groups. then the

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Current Student
User avatar
Joined: 11 May 2008
Posts: 561
Followers: 7

Kudos [?]: 15 [0], given: 0

GMAT Tests User
six students are equally divided into 3 groups. then the [#permalink] New post 11 Aug 2008, 00:31
six students are equally divided into 3 groups. then the three groups were assigned to three different topics. how many diff arrangements are there?
30
60
90
180
540
it seems simple, but i could not get the ans...
i thought ...
ways of selecting 2 students out of 6 is 6c2
and each grp has 3 topics.
so no. of poss arrangements = 6c2*3 .
where have i gone wrong??
Senior Manager
Senior Manager
Joined: 06 Apr 2008
Posts: 452
Followers: 1

Kudos [?]: 32 [0], given: 1

Re: combination [#permalink] New post 11 Aug 2008, 00:42
arjtryarjtry wrote:
six students are equally divided into 3 groups. then the three groups were assigned to three different topics. how many diff arrangements are there?
30
60
90
180
540
it seems simple, but i could not get the ans...
i thought ...
ways of selecting 2 students out of 6 is 6c2
and each grp has 3 topics.
so no. of poss arrangements = 6c2*3 .
where have i gone wrong??


Ways of selecting group = 6C2 * 4C2 * 2C2 / 3! = 15

Three groups can select three subjects in 6 ways

Therefore total combinations = 15*6 = 90
Manager
Manager
Joined: 15 Jul 2008
Posts: 210
Followers: 3

Kudos [?]: 17 [0], given: 0

GMAT Tests User
Re: combination [#permalink] New post 11 Aug 2008, 03:57
Number of ways of forming 3 teams = 6C2*4C2*3C2/3! = 15
Number of ways of assigning the 3 teams to 3 tasks = 3!
Number of ways of performing the 3 tasks = 3!

So total arrangements = 15*3!*3! = 540
Director
Director
Joined: 27 May 2008
Posts: 552
Followers: 5

Kudos [?]: 143 [0], given: 0

GMAT Tests User
Re: combination [#permalink] New post 11 Aug 2008, 09:11
bhushangiri wrote:
Number of ways of forming 3 teams = 6C2*4C2*3C2/3! = 15
Number of ways of assigning the 3 teams to 3 tasks = 3!
Number of ways of performing the 3 tasks = 3!

So total arrangements = 15*3!*3! = 540


could you explain the highlighted step... i'm getting 90 = 15 * 3!

suppose the students are numbered 1,2,3,4,5,6 and tasks are X,Y and Z

one of the 15 possible ways of forming teams is 12, 34, 56. these teams can be assigned to 3 tasks in 3! = 6 ways
X-- Y-- Z
12-- 34-- 56
12-- 56-- 34
34-- 12-- 56
34-- 56-- 12
56-- 12-- 34
56-- 34-- 12
so the answer should be 15*6 = 90
Manager
Manager
Joined: 15 Jul 2008
Posts: 210
Followers: 3

Kudos [?]: 17 [0], given: 0

GMAT Tests User
Re: combination [#permalink] New post 11 Aug 2008, 09:18
90 is the number of ways you can assign 3 teams formed out of 12 people to 3 different tasks.
But now you can order the 3 tasks in 3! ways. T1 T2 T3 or T2 T1 T3.... etc etc.

I was confused between 90 and 540 but since question used the word "arrangements" decided to go with complete arrangements including the order of tasks.

durgesh79 wrote:
bhushangiri wrote:
Number of ways of forming 3 teams = 6C2*4C2*3C2/3! = 15
Number of ways of assigning the 3 teams to 3 tasks = 3!
Number of ways of performing the 3 tasks = 3!

So total arrangements = 15*3!*3! = 540


could you explain the highlighted step... i'm getting 90 = 15 * 3!

suppose the students are numbered 1,2,3,4,5,6 and tasks are X,Y and Z

one of the 15 possible ways of forming teams is 12, 34, 56. these teams can be assigned to 3 tasks in 3! = 6 ways
X-- Y-- Z
12-- 34-- 56
12-- 56-- 34
34-- 12-- 56
34-- 56-- 12
56-- 12-- 34
56-- 34-- 12
so the answer should be 15*6 = 90

But now you can fruther decide which task you want to perform first X Y or Z..


Last edited by bhushangiri on 11 Aug 2008, 09:20, edited 1 time in total.
Senior Manager
Senior Manager
Joined: 06 Apr 2008
Posts: 452
Followers: 1

Kudos [?]: 32 [0], given: 1

Re: combination [#permalink] New post 11 Aug 2008, 09:19
durgesh79 wrote:
bhushangiri wrote:
Number of ways of forming 3 teams = 6C2*4C2*3C2/3! = 15
Number of ways of assigning the 3 teams to 3 tasks = 3!
Number of ways of performing the 3 tasks = 3!

So total arrangements = 15*3!*3! = 540


could you explain the highlighted step... i'm getting 90 = 15 * 3!

suppose the students are numbered 1,2,3,4,5,6 and tasks are X,Y and Z

one of the 15 possible ways of forming teams is 12, 34, 56. these teams can be assigned to 3 tasks in 3! = 6 ways
X-- Y-- Z
12-- 34-- 56
12-- 56-- 34
34-- 12-- 56
34-- 56-- 12
56-- 12-- 34
56-- 34-- 12
so the answer should be 15*6 = 90


You also need to consider if tasks assigned are X-12, Y-34, Z-56 then these tasks can be performed in following order

X-12 Y-34 Z-56
X-12 Y-56 Z-34
Y-34 X-12 Z-56
Y-34 Z-56 X-12
Z-56 Y-34 X-12
Z-56 X-12 Y-34

Therefore total combinations = 15*6*6
SVP
SVP
User avatar
Joined: 30 Apr 2008
Posts: 1893
Location: Oklahoma City
Schools: Hard Knocks
Followers: 27

Kudos [?]: 398 [0], given: 32

GMAT Tests User
Re: combination [#permalink] New post 11 Aug 2008, 09:26
Yes, the answer is E

C_6^2 * C_4^2 * P_3
_________________

------------------------------------
J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

Director
Director
Joined: 27 May 2008
Posts: 552
Followers: 5

Kudos [?]: 143 [0], given: 0

GMAT Tests User
Re: combination [#permalink] New post 11 Aug 2008, 10:02
bhushangiri wrote:
But now you can fruther decide which task you want to perform first X Y or Z..


nmohindru wrote:
You also need to consider if tasks assigned are X-12, Y-34, Z-56 then these tasks can be performed in following order


i'd disagree, i dont think order of task is important here .... how do we know that 3 teams cant perform 3 tasks simultaneously....

if i "have to" arrive at 540... i'll use following logic .. :P

ways of arranging of 6 students in 3 teams = 6!/2!*2!*2! ( arranging 6 things in row, with 2 each in 3 groups)
= 90

each team can be assigned taskes in 3! ways

total arrangements = 90 * 3! = 540.

but i still dont agree with this answer :? can someone please explain..
whats the source and OE.
Manager
Manager
Joined: 15 Jul 2008
Posts: 210
Followers: 3

Kudos [?]: 17 [0], given: 0

GMAT Tests User
Re: combination [#permalink] New post 11 Aug 2008, 10:23
durgesh79 wrote:
bhushangiri wrote:
But now you can fruther decide which task you want to perform first X Y or Z..


nmohindru wrote:
You also need to consider if tasks assigned are X-12, Y-34, Z-56 then these tasks can be performed in following order


i'd disagree, i dont think order of task is important here .... how do we know that 3 teams cant perform 3 tasks simultaneously....

if i "have to" arrive at 540... i'll use following logic .. :P

ways of arranging of 6 students in 3 teams = 6!/2!*2!*2! ( arranging 6 things in row, with 2 each in 3 groups)
= 90

each team can be assigned taskes in 3! ways

total arrangements = 90 * 3! = 540.

but i still dont agree with this answer :? can someone please explain..
whats the source and OE.


Which is a fair argument. But since both the options were given, and the question asked "arrangements were possible", i chose 540.
Re: combination   [#permalink] 11 Aug 2008, 10:23
    Similar topics Author Replies Last post
Similar
Topics:
New posts 4 Experts publish their posts in the topic A group of n students can be divided into equal groups of 4 alimad 9 03 Nov 2007, 04:32
New posts The 52 students in a certain class are divided to 7 groups. arjtryarjtry 5 28 Jul 2008, 19:09
New posts 1 If 72 cupcakes must be divided equally among the students in manalq8 5 19 Nov 2011, 11:31
Popular new posts 2 Experts publish their posts in the topic A group of n students can be divided into equal groups of 4 enigma123 11 21 Jan 2012, 13:37
New posts Experts publish their posts in the topic A class is divided into four groups of four students each. docabuzar 2 02 Feb 2012, 11:07
Display posts from previous: Sort by

six students are equally divided into 3 groups. then the

  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®.