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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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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
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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 X12, Y34, Z56 then these tasks can be performed in following order X12 Y34 Z56 X12 Y56 Z34 Y34 X12 Z56 Y34 Z56 X12 Z56 Y34 X12 Z56 X12 Y34 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
Yes, the answer is E \(C_6^2 * C_4^2 * P_3\)
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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 X12, Y34, Z56 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 X12, Y34, Z56 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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Re: Six students are equally divided into 3 groups, then, the three groups [#permalink]
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26 May 2017, 20:51
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 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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27 May 2017, 02:04
s1, (s2,s3,s4,s5,s6)  G1 s2, (s3,s4,s5,s6)  G2 s3, (s4,s5,s6)  G3 s4, (s5,s6)  G4 s5, (s6)  G5
Total groups 15.
3 topics A,B,C can be distributed in 6 (3!) different ways to 15 different groups. Hence 15*6=90, arrangements are possible.



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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, 06:52
SravnaTestPrepWhy are we dividing the 6c2* 4c2 * 2c2 by 3!?



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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, 07:07
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Nitish7 wrote: SravnaTestPrepWhy are we dividing the 6c2* 4c2 * 2c2 by 3!? We are dividing by 3! because the order of groups is not important Take a simple example of 4 people A,B,C,D being divided into 2 equal groups The groups may be: (A,B) and (C,D) (A,C) and (B,D) (A,D) and (B,C) When we are computing the possibilities using 4C2*2c2 we are also including (C,D) and (AB) (B,D) and (A,C) (B,C) and (A,D) But we know the latter need not be counted because (A,B) and (C,D) is the same as (C,D) and (A,B) To get the required number of possibilities without repeating, we divide by (no.of groups)!
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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, 07:41
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, 10:43
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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!
Answer: 6C2 * 3! = 90
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Re: Six students are equally divided into 3 groups, then, the three groups [#permalink]
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09 Jun 2017, 02:27
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?



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Re: Six students are equally divided into 3 groups, then, the three groups [#permalink]
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09 Jun 2017, 02:47
Nitish7 wrote: 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? 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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Re: Six students are equally divided into 3 groups, then, the three groups [#permalink]
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09 Jun 2017, 08:35
Imo C 3 groups can be formed in 15 ways then 3 topics can be chosen in 6 ways So number of arrangements =90 Sent from my ONE E1003 using GMAT Club Forum mobile app




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