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 350,000 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:

six students are equally divided into 3 groups. then the [#permalink]
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??

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

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!

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

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.

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

Great to know you are joining Kellogg. A lot was being talked about your last minute interview on Pagalguy (all good though). It was kinda surprise that you got the...