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Current Student
Joined: 11 May 2008
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six students are equally divided into 3 groups. then the [#permalink]
 11 Aug 2008, 01:31
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??
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Senior Manager
Joined: 06 Apr 2008
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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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Manager
Joined: 15 Jul 2008
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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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Director
Joined: 27 May 2008
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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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Manager
Joined: 15 Jul 2008
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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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Senior Manager
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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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SVP
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Yes, the answer is E C_6^2 * C_4^2 * P_3
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Director
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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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Manager
Joined: 15 Jul 2008
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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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