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

 It is currently 02 Jul 2015, 15:19

### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# six students are equally divided into 3 groups. then the

Author Message
TAGS:
Current Student
Joined: 11 May 2008
Posts: 560
Followers: 7

Kudos [?]: 59 [0], given: 0

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??
Senior Manager
Joined: 06 Apr 2008
Posts: 449
Followers: 1

Kudos [?]: 71 [0], given: 1

Re: combination [#permalink]  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
Joined: 15 Jul 2008
Posts: 208
Followers: 3

Kudos [?]: 31 [0], given: 0

Re: combination [#permalink]  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
Joined: 27 May 2008
Posts: 550
Followers: 7

Kudos [?]: 223 [0], given: 0

Re: combination [#permalink]  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
Joined: 15 Jul 2008
Posts: 208
Followers: 3

Kudos [?]: 31 [0], given: 0

Re: combination [#permalink]  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
Joined: 06 Apr 2008
Posts: 449
Followers: 1

Kudos [?]: 71 [0], given: 1

Re: combination [#permalink]  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
Joined: 30 Apr 2008
Posts: 1889
Location: Oklahoma City
Schools: Hard Knocks
Followers: 35

Kudos [?]: 473 [0], given: 32

Re: combination [#permalink]  11 Aug 2008, 09:26

$$C_6^2 * C_4^2 * P_3$$
_________________

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

GMAT Club Premium Membership - big benefits and savings

Director
Joined: 27 May 2008
Posts: 550
Followers: 7

Kudos [?]: 223 [0], given: 0

Re: combination [#permalink]  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 ..

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
Joined: 15 Jul 2008
Posts: 208
Followers: 3

Kudos [?]: 31 [0], given: 0

Re: combination [#permalink]  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 ..

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 Replies Last post
Similar
Topics:
People in a line are divided into groups. Sam was the 27th 1 10 Oct 2008, 20:26
The 52 students in a certain class are divided to 7 groups. 5 28 Jul 2008, 19:09
There are 5 people in group A, 4 in group B, and 3 in group 2 05 May 2006, 23:23
If 6 people are to be divided to 3 different groups, each of 8 05 May 2006, 22:06
When S is divided by 5 remainder is 3, when it is divided by 16 26 Apr 2006, 02:32
Display posts from previous: Sort by