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

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11 Aug 2008, 00:31

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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?

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

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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

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

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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!

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

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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

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

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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.

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

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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

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

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12 Jul 2015, 06:54

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

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25 Jul 2016, 23:04

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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

1. Take a simple example of 4 students into 2 groups of 2 students each and two different topics being assigned . 2. Taking an example case let the students be s1,s2,s3,s4. the groups could be (s1,s2) and (s3,s4), (s1,s3) and (s2,s4) , (s1,s4) and (s2,s3), Number of ways of forming groups is 4C2*2C2/2!=3 where 2 in the denominator represents the number of groups. Remember order of groups is not important. 3. Applying the above rule to the current case, number of ways of forming groups is 6C2*4C2/3!. 4. Three topics can be arranged between the groups in 3! ways. 5. Total number of possible arrangements is (3)*(4)=90
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Re: Six students are equally divided into 3 groups, then, the three groups [#permalink]

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08 Jun 2017, 06:41

arjtryarjtry wrote:

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

The question is really tough.. I got 540 as answer but after reading the solution understood why the no. of different arrangements should be 90 only..
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Re: Six students are equally divided into 3 groups, then, the three groups [#permalink]

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08 Jun 2017, 09:43

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Rephrase the question: How many 2 person arrangements can you make out of 6 people = 6C2= 15. Then how many ways can you assign the 2 person teams to 3 assignments= 3!

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

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09 Jun 2017, 01:27

Hi,

Thanks for your valuable feedback please let me know one more thing as the question asks us to create 3 equal groups from 6 students so in each group there will be 2 students so if I use 6C2 will it be correct?

Thanks for your valuable feedback please let me know one more thing as the question asks us to create 3 equal groups from 6 students so in each group there will be 2 students so if I use 6C2 will it be correct?

It is the number of ways of first selecting 2 students out of 6 and then 2 out of the remaining 4 and then 2 out of the remaining 2. So it is 6C2*4C2*2C2. . But since order is not important, as I explained in my previous post, 6C2*4C2*2C2 has to be divided by (no.of groups)!
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