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Six students are equally divided into 3 groups, then, the three groups [#permalink]
11 Aug 2008, 00:31

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Difficulty:

65% (hard)

Question Stats:

42% (02:09) correct
58% (01:17) wrong based on 33 sessions

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]
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]
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]
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]
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]
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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