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# Six students are equally divided into 3 groups, then, the three groups

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Six students are equally divided into 3 groups, then, the three groups [#permalink]

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11 Aug 2008, 01:31
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Question Stats:

43% (02:10) correct 57% (01:34) wrong based on 108 sessions

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Six students are equally divided into 3 groups, then, the three groups were assigned to three different topics. How many different arrangements are possible?

(A) 30
(B) 60
(C) 90
(D) 180
(E) 540
[Reveal] Spoiler: OA

Last edited by Bunuel on 13 Jul 2015, 00:37, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.
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Re: Six students are equally divided into 3 groups, then, the three groups [#permalink]

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11 Aug 2008, 01:42
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
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Re: Six students are equally divided into 3 groups, then, the three groups [#permalink]

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11 Aug 2008, 04:57
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
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Re: Six students are equally divided into 3 groups, then, the three groups [#permalink]

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11 Aug 2008, 10:11
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
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Re: Six students are equally divided into 3 groups, then, the three groups [#permalink]

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11 Aug 2008, 10:18
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, 10:20, edited 1 time in total.
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Re: Six students are equally divided into 3 groups, then, the three groups [#permalink]

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11 Aug 2008, 10:19
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
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Re: Six students are equally divided into 3 groups, then, the three groups [#permalink]

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11 Aug 2008, 10:26

$$C_6^2 * C_4^2 * P_3$$
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Re: Six students are equally divided into 3 groups, then, the three groups [#permalink]

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11 Aug 2008, 11:02
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.
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Re: Six students are equally divided into 3 groups, then, the three groups [#permalink]

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11 Aug 2008, 11:23
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.
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Re: Six students are equally divided into 3 groups, then, the three groups [#permalink]

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12 Jul 2015, 07:54
Hello from the GMAT Club BumpBot!

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Re: Six students are equally divided into 3 groups, then, the three groups [#permalink]

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26 Jul 2016, 00:04
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).

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Re: Six students are equally divided into 3 groups, then, the three groups   [#permalink] 26 Jul 2016, 00:04
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