Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 05 Jun 2010
Posts: 3

6 people form groups of 2 for a practical work. Each group [#permalink]
Show Tags
05 Jun 2010, 15:53
Question Stats:
60% (01:01) correct 40% (00:46) wrong based on 27 sessions
HideShow timer Statistics
A) 6 people form groups of 2 for a practical work. Each group is assigned one of three continents: Asia, Europe or Africa. In how many different ways can the work be organized?
B) In a group of 10 people, 6 women and 4 men. If a comission of three people has to be formed with at least one man, how many groups can we form?



Math Expert
Joined: 02 Sep 2009
Posts: 46129

Re: Combinations problems [#permalink]
Show Tags
05 Jun 2010, 16:21
Gusano97 wrote: A) 6 people form groups of 2 for a practical work. Each group is assigned one of three continents: Asia, Europe or Africa. In how many different ways can the work be organized?
B) In a group of 10 people, 6 women and 4 men. If a comission of three people has to be formed with at least one man, how many groups can we form? Hi, and welcome to the Gmat Club. Below are the solutions for your problems. Hope it helps. A. 6 people form groups of 2 for a practical work. Each group is assigned one of three continents: Asia, Europe or Africa. In how many different ways can the work be organized?# of ways 6 people can be divided into 3 groups when order matters is: \(C^2_6*C^2_4*C^2_2=90\). Answer: 90. Similar topics: probability85993.html?highlight=divide+groupscombination55369.html#p690842probability88685.html#p669025combinationgroupsandthatstuff85707.html#p642634subcommittee86346.html?highlight=divide+groupsB. In a group of 10 people, 6 women and 4 men. If a comission of three people has to be formed with at least one man, how many groups can we form?Let's find the probability of the opposite event and subtract it from 1. Opposite event would be that in the committee of 3 won't be any man (so only women)  \(P(m=0)=P(w=3)=\frac{C^3_6}{C^3_{10}}=\frac{1}{6}\). \(C^3_6\)  # of ways to choose 3 women out 6 women; \(C^3_{10}\)  total # of ways to choose 3 people out of 10. \(P(m\geq{1})=1P(m=0)=1\frac{1}{6}=\frac{5}{6}\). Answer: \(\frac{5}{6}\)
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 05 Jun 2010
Posts: 3

Re: Combinations problems [#permalink]
Show Tags
05 Jun 2010, 16:25
Thank you Bunuel!.



Manager
Joined: 14 Apr 2010
Posts: 196

Re: Combinations problems [#permalink]
Show Tags
07 Jun 2010, 06:35
A. 6 people form groups of 2 for a practical work. Each group is assigned one of three continents: Asia, Europe or Africa. In how many different ways can the work be organized? # of ways 6 people can be divided into 3 groups when order matters is: . Answer: 90. Bunuel, I have always found your explanations brilliant. With this question, however, i could not grasp your explanation. Please kindly elaborate....



Senior Manager
Joined: 25 Jun 2009
Posts: 287

Re: Combinations problems [#permalink]
Show Tags
07 Jun 2010, 06:53
Bunuel wrote: Gusano97 wrote: A) 6 people form groups of 2 for a practical work. Each group is assigned one of three continents: Asia, Europe or Africa. In how many different ways can the work be organized?
B) In a group of 10 people, 6 women and 4 men. If a comission of three people has to be formed with at least one man, how many groups can we form? Hi, and welcome to the Gmat Club. Below are the solutions for your problems. Hope it helps. A. 6 people form groups of 2 for a practical work. Each group is assigned one of three continents: Asia, Europe or Africa. In how many different ways can the work be organized?# of ways 6 people can be divided into 3 groups when order matters is: \(C^2_6*C^2_4*C^2_2=90\). Answer: 90. Similar topics: probability85993.html?highlight=divide+groupscombination55369.html#p690842probability88685.html#p669025combinationgroupsandthatstuff85707.html#p642634subcommittee86346.html?highlight=divide+groupsBunuel, Dont you think we need to multiply 90 with 3! as we have 3 teams now and 3 teams needs to be given one place amongst asia, europe or Africa, Please correct me if my understanding is wrong here.



Math Expert
Joined: 02 Sep 2009
Posts: 46129

Re: Combinations problems [#permalink]
Show Tags
07 Jun 2010, 08:23
bibha wrote: A. 6 people form groups of 2 for a practical work. Each group is assigned one of three continents: Asia, Europe or Africa. In how many different ways can the work be organized? # of ways 6 people can be divided into 3 groups when order matters is: . Answer: 90. Bunuel, I have always found your explanations brilliant. With this question, however, i could not grasp your explanation. Please kindly elaborate.... nitishmahajan wrote: Bunuel,
Dont you think we need to multiply 90 with 3! as we have 3 teams now and 3 teams needs to be given one place amongst asia, europe or Africa,
Please correct me if my understanding is wrong here. There are 15 ways 6 people can be divided equally into 3 groups, each containing 2 persons when the order of the groups is not important (meaning that we don't have group #1, group #2, and group #3): 1. {AB}{CD}{EF} 2. {AB}{CE}{DF} 3. {AB}{CF}{DE} 4. {AC}{BD}{EF} 5. {AC}{BE}{DF} 6. {AC}{BF}{DE} 7. {AD}{BC}{EF} 8. {AD}{BE}{CF} 9. {AD}{BF}{CE} 10. {AE}{CD}{BF} 11. {AE}{BC}{DF} 12. {AE}{CF}{BD} 13. {AF}{CD}{BE} 14. {AF}{BC}{DE} 15. {AF}{CE}{BD} We can get this # from the following formula: \(\frac{C^2_6*C^2_4*C^2_2}{3!}=15\). We are dividing by 3! (factorial of the # of groups) because the order of the groups is not important. Next, we are told that each group is assigned to one of three continents: Asia, Europe or Africa. So now the order of the groups IS important as we can assign group #1 to Asia, #2 to Europe and #3 to Africa OR #1 to Europe, #2 to Asia and #3 to Africa ... Several different assignments are possible. So how many? Let's consider division #1 (1. {AB}{CD}{EF}), in how many ways can we assign these groups to the given 3 countries? Asia  Europe  Africa{AB}  {CD}  {EF} {AB}  {EF}  {CD} {CD}  {AB}  {EF} {CD}  {EF}  {AB} {EF}  {AB}  {CD} {EF}  {CD}  {AB} Total 6 different assignments, basically the # of permutations of 3 distinct objects (3!): {AB}, {CD}, and {EF}. 3!=6 different assignments for one particular division, means that for 15 divisions there will be total of 3!*15=90 assignments possible. So when the the order of the groups is not important we are dividing \(C^2_6*C^2_4*C^2_2\) by the factorial of the # of groups  3! > \(\frac{C^2_6*C^2_4*C^2_2}{3!}=15\); But when the order of the groups is important (when we are assigning them to certain task) divisionj is not deeded: \(C^2_6*C^2_4*C^2_2=90\). Pleas check the following links for similar problems: probability85993.html?highlight=divide+groupscombination55369.html#p690842probability88685.html#p669025combinationgroupsandthatstuff85707.html#p642634subcommittee86346.html?highlight=divide+groupsHope it's clear.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Senior Manager
Joined: 25 Jun 2009
Posts: 287

Re: Combinations problems [#permalink]
Show Tags
07 Jun 2010, 08:28
Bunuel wrote: bibha wrote: A. 6 people form groups of 2 for a practical work. Each group is assigned one of three continents: Asia, Europe or Africa. In how many different ways can the work be organized? # of ways 6 people can be divided into 3 groups when order matters is: . Answer: 90. Bunuel, I have always found your explanations brilliant. With this question, however, i could not grasp your explanation. Please kindly elaborate.... nitishmahajan wrote: Bunuel,
Dont you think we need to multiply 90 with 3! as we have 3 teams now and 3 teams needs to be given one place amongst asia, europe or Africa,
Please correct me if my understanding is wrong here. There are 15 ways 6 people can be divided equally into 3 groups, each containing 2 persons when the order of the groups is not important (meaning that we don't have group #1, group #2, and group #3): 1. {AB}{CD}{EF} 2. {AB}{CE}{DF} 3. {AB}{CF}{DE} 4. {AC}{BD}{EF} 5. {AC}{BE}{DF} 6. {AC}{BF}{DE} 7. {AD}{BC}{EF} 8. {AD}{BE}{CF} 9. {AD}{BF}{CE} 10. {AE}{CD}{BF} 11. {AE}{BC}{DF} 12. {AE}{CF}{BD} 13. {AF}{CD}{BE} 14. {AF}{BC}{DE} 15. {AF}{CE}{BD} We can get this # from the following formula: \(\frac{C^2_6*C^2_4*C^2_2}{3!}=15\). We are dividing by 3! (factorial of the # of groups) because the order of the groups is not important. Next, we are told that each group is assigned to one of three continents: Asia, Europe or Africa. So now the order of the groups IS important as we can assign group #1 to Asia, #2 to Europe and #3 to Africa OR #1 to Europe, #2 to Asia and #3 to Africa ... Several different assignments are possible. So how many? Let's consider division #1 (1. {AB}{CD}{EF}), in how many ways can we assign these groups to the given 3 countries? Asia  Europe  Africa{AB}  {CD}  {EF} {AB}  {EF}  {CD} {CD}  {AB}  {EF} {CD}  {EF}  {AB} {EF}  {AB}  {CD} {EF}  {CD}  {AB} Total 6 different assignments, basically the # of permutations of 3 distinct objects (3!): {AB}, {CD}, and {EF}. 3!=6 different assignments for one particular division, means that for 15 divisions there will be total of 3!*15=90 assignments possible. So when the the order of the groups is not important we are dividing \(C^2_6*C^2_4*C^2_2\) by the factorial of the # of groups  3! > \(\frac{C^2_6*C^2_4*C^2_2}{3!}=15\); But when the order of the groups is important (when we are assigning them to certain task) divisionj is not deeded: \(C^2_6*C^2_4*C^2_2=90\). Pleas check the following links for similar problems: probability85993.html?highlight=divide+groupscombination55369.html#p690842probability88685.html#p669025combinationgroupsandthatstuff85707.html#p642634subcommittee86346.html?highlight=divide+groupsHope it's clear. Thanks Bunuel, Its clear now ..!



Intern
Joined: 21 Jun 2010
Posts: 1

Re: Combinations problems [#permalink]
Show Tags
21 Jun 2010, 07:24
Bunuel,
I'm confused on the second question. Doesn't it ask for a specific number, not a probability?
In any event, would it by 5/6ths of the total number > 5/6ths of 120 > 100?
Thanks for your help.



Math Expert
Joined: 02 Sep 2009
Posts: 46129

Re: Combinations problems [#permalink]
Show Tags
21 Jun 2010, 07:35



Intern
Joined: 22 Jun 2010
Posts: 6

Re: Combinations problems [#permalink]
Show Tags
23 Jun 2010, 23:56
buneul, just one more favor....
please let me know if my approach is correct.
first i calculated the number of ways in which at least one man is chosen
1. MWW 4 2. MMW 6 3 MMM 4
then for each i calculated the number of ways the women can be filled in... so for 1. 6!/2!(4!) = 15 2. 6!/5! = 6 3. none
then 4(15) + 6(6) + 4 = 100



Math Expert
Joined: 02 Sep 2009
Posts: 46129

Re: Combinations problems [#permalink]
Show Tags
24 Jun 2010, 05:34



Senior Manager
Joined: 29 Sep 2009
Posts: 383

Re: Combinations problems [#permalink]
Show Tags
23 Oct 2010, 20:11
The explanation is clear, but I have a few doubts: 1. We define nCr as r selections out of n when the order is not important. If the order is important we use nPr. 2. 6C2 (2 out of 6)x 4C2 (2 out of 4) x 2C2(2 out of 2), now since "C" by definition is used for selections when order is not important  why divide? The answer is right, but I cant understand how we arrived at the formula/division.



Math Expert
Joined: 02 Sep 2009
Posts: 46129

Re: Combinations problems [#permalink]
Show Tags
24 Oct 2010, 03:50



Senior Manager
Joined: 29 Sep 2009
Posts: 383

Re: Combinations problems [#permalink]
Show Tags
24 Oct 2010, 05:25
Bunuel wrote: vicksikand wrote: The explanation is clear, but I have a few doubts: 1. We define nCr as r selections out of n when the order is not important. If the order is important we use nPr. 2. 6C2 (2 out of 6)x 4C2 (2 out of 4) x 2C2(2 out of 2), now since "C" by definition is used for selections when order is not important  why divide? The answer is right, but I cant understand how we arrived at the formula/division. Read this post: combinationsproblems95344.html#p734396 Similar problems about the same concept: probability85993.html?highlight=divide+groupscombination55369.html#p690842probability88685.html#p669025combinationgroupsandthatstuff85707.html#p642634subcommittee86346.html?highlight=divide+groupsHope it helps. Thats the explanation I was talking about. I read it  it totally makes sense , but how did you arrive at the formula? "We can get this # from the following formula: \frac{C^2_6*C^2_4*C^2_2}{3!}=15. We are dividing by 3! (factorial of the # of groups) because the order of the groups is not important." nCr is used for selections when order isnt important, nPr when order is important.



Math Expert
Joined: 02 Sep 2009
Posts: 46129

Re: Combinations problems [#permalink]
Show Tags
24 Oct 2010, 05:28



Senior Manager
Joined: 29 Sep 2009
Posts: 383

Re: Combinations problems [#permalink]
Show Tags
24 Oct 2010, 05:43
Hope it helps.[/quote] Thats the explanation I was talking about. I read it  it totally makes sense , but how did you arrive at the formula? "We can get this # from the following formula: \frac{C^2_6*C^2_4*C^2_2}{3!}=15. We are dividing by 3! (factorial of the # of groups) because the order of the groups is not important." nCr is used for selections when order isnt important, nPr when order is important.[/quote] combination55369.html#p690685[/quote] clear ! Thanks



Manager
Joined: 03 Aug 2010
Posts: 101
GMAT Date: 08082011

Re: Combinations problems [#permalink]
Show Tags
31 Oct 2010, 09:25
Bunuel wrote: Gusano97 wrote: A) 6 people form groups of 2 for a practical work. Each group is assigned one of three continents: Asia, Europe or Africa. In how many different ways can the work be organized?
B) In a group of 10 people, 6 women and 4 men. If a comission of three people has to be formed with at least one man, how many groups can we form? Hi, and welcome to the Gmat Club. Below are the solutions for your problems. Hope it helps. A. 6 people form groups of 2 for a practical work. Each group is assigned one of three continents: Asia, Europe or Africa. In how many different ways can the work be organized?# of ways 6 people can be divided into 3 groups when order matters is: \(C^2_6*C^2_4*C^2_2=90\). Answer: 90. Bunuel, why does order matter in the question above? In the question below, the answer would be 2520 if order matters, but the answer is 105. Are these two questions not essentially the same? In how many different ways can a group of 8 people be divided into 4 teams of 2 people each? (A) 90 (B) 105 (C) 168 (D) 420 (E) 2520



Math Expert
Joined: 02 Sep 2009
Posts: 46129

Re: Combinations problems [#permalink]
Show Tags
31 Oct 2010, 09:40
Yalephd wrote: Bunuel wrote: Gusano97 wrote: A) 6 people form groups of 2 for a practical work. Each group is assigned one of three continents: Asia, Europe or Africa. In how many different ways can the work be organized?
B) In a group of 10 people, 6 women and 4 men. If a comission of three people has to be formed with at least one man, how many groups can we form? Hi, and welcome to the Gmat Club. Below are the solutions for your problems. Hope it helps. A. 6 people form groups of 2 for a practical work. Each group is assigned one of three continents: Asia, Europe or Africa. In how many different ways can the work be organized?# of ways 6 people can be divided into 3 groups when order matters is: \(C^2_6*C^2_4*C^2_2=90\). Answer: 90. Bunuel, why does order matter in the question above? In the question below, the answer would be 2520 if order matters, but the answer is 105. Are these two questions not essentially the same? In how many different ways can a group of 8 people be divided into 4 teams of 2 people each? (A) 90 (B) 105 (C) 168 (D) 420 (E) 2520 Whole above discussion is about this issue. Please read it and follow the links provided in my earlier posts for other examples and discussions of this concept. But again: Assignment: Asia  group#1, Europe  group#2, Africa  group#3 is different from Asia  group#3, Europe  group#1, Africa  group#2. That's why the order of the groups matters here. Second problem discussed here: combinationgroupsandthatstuff85707.html#p642634In this problem there is no group#1, group#2, group#3, group#4 so 2520 should be divided by 4!=105 to get rid of duplications.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Senior Manager
Joined: 24 Aug 2009
Posts: 481
Schools: Harvard, Columbia, Stern, Booth, LSB,

Re: Combinations problems [#permalink]
Show Tags
16 Sep 2012, 07:10
Bunuel wrote: Gusano97 wrote: A) 6 people form groups of 2 for a practical work. Each group is assigned one of three continents: Asia, Europe or Africa. In how many different ways can the work be organized?
B) In a group of 10 people, 6 women and 4 men. If a comission of three people has to be formed with at least one man, how many groups can we form? Hi, and welcome to the Gmat Club. Below are the solutions for your problems. Hope it helps. A. 6 people form groups of 2 for a practical work. Each group is assigned one of three continents: Asia, Europe or Africa. In how many different ways can the work be organized?# of ways 6 people can be divided into 3 groups when order matters is: \(C^2_6*C^2_4*C^2_2=90\). Answer: 90. Hi Bunuel, In reference to the first question, i have a doubt which is 90 is the no of ways in which 6 people can be divided into 3 groups of 2 persons each. Shouldn't the answer be 90 x 6 = 540 because these 3 different can be sent to 3 different location in 3! ways. Kindly correct me if i am wrong. Waiting for your reply.
_________________
If you like my Question/Explanation or the contribution, Kindly appreciate by pressing KUDOS. Kudos always maximizes GMATCLUB worth Game Theory
If you have any question regarding my post, kindly pm me or else I won't be able to reply



Senior Manager
Joined: 24 Aug 2009
Posts: 481
Schools: Harvard, Columbia, Stern, Booth, LSB,

Re: Combinations problems [#permalink]
Show Tags
16 Sep 2012, 07:10
Bunuel wrote: Gusano97 wrote: A) 6 people form groups of 2 for a practical work. Each group is assigned one of three continents: Asia, Europe or Africa. In how many different ways can the work be organized?
B) In a group of 10 people, 6 women and 4 men. If a comission of three people has to be formed with at least one man, how many groups can we form? Hi, and welcome to the Gmat Club. Below are the solutions for your problems. Hope it helps. A. 6 people form groups of 2 for a practical work. Each group is assigned one of three continents: Asia, Europe or Africa. In how many different ways can the work be organized?# of ways 6 people can be divided into 3 groups when order matters is: \(C^2_6*C^2_4*C^2_2=90\). Answer: 90. Hi Bunuel, In reference to the first question, i have a doubt which is 90 is the no of ways in which 6 people can be divided into 3 groups of 2 persons each. Shouldn't the answer be 90 x 6 = 540 because these 3 different teams can be sent to 3 different location in 3! ways. Kindly correct me if i am wrong. Waiting for your reply.
_________________
If you like my Question/Explanation or the contribution, Kindly appreciate by pressing KUDOS. Kudos always maximizes GMATCLUB worth Game Theory
If you have any question regarding my post, kindly pm me or else I won't be able to reply




Re: Combinations problems
[#permalink]
16 Sep 2012, 07:10



Go to page
1 2
Next
[ 27 posts ]



