Author 
Message 
TAGS:

Hide Tags

Manager
Status: Keep fighting!
Affiliations: IIT Madras
Joined: 31 Jul 2010
Posts: 206
WE 1: 2+ years  Programming
WE 2: 3+ years  Product developement,
WE 3: 2+ years  Program management

In how many different ways can a group of 9 people be divide
[#permalink]
Show Tags
26 Sep 2010, 09:47
Question Stats:
51% (00:54) correct 49% (01:10) wrong based on 611 sessions
HideShow timer Statistics
In how many different ways can a group of 9 people be divided into 3 groups, with each group containing 3 people? A. 280 B. 1,260 C. 1,680 D. 2,520 E. 3,360
Official Answer and Stats are available only to registered users. Register/ Login.




Math Expert
Joined: 02 Sep 2009
Posts: 49303

Re: 9 people and Combinatorics
[#permalink]
Show Tags
26 Sep 2010, 09:56
hemanthp wrote: In how many different ways can a group of 9 people be divided into 3 groups, with each group containing 3 people?
280 1,260 1,680 2,520 3,360 GENERAL RULE:1. The number of ways in which \(mn\) different items can be divided equally into \(m\) groups, each containing \(n\) objects and the order of the groups is important is \(\frac{(mn)!}{(n!)^m}\) 2. The number of ways in which \(mn\) different items can be divided equally into \(m\) groups, each containing \(n\) objects and the order of the groups is NOT important is \(\frac{(mn)!}{(n!)^m*m!}\). BACK TO THE ORIGINAL QUESTION:In original question I think the order is NOT important, as we won't have group #1, #2 and #3. So we should use second formula, \(mn=9\), \(m=3\) groups \(n=3\) objects (people): \(\frac{(mn)!}{(n!)^m*m!}=\frac{9!}{(3!)^3*3!}=280\). This can be done in another way as well: \(\frac{9C3*6C3*3C3}{3!}=280\), we are dividing by \(3!\) as there are 3 groups and order doesn't matter. Answer: A.
_________________
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




Retired Moderator
Joined: 02 Sep 2010
Posts: 772
Location: London

Re: 9 people and Combinatorics
[#permalink]
Show Tags
02 Oct 2010, 00:41
hemanthp wrote: In how many different ways can a group of 9 people be divided into 3 groups, with each group containing 3 people?
280 1,260 1,680 2,520 3,360 To divide 9 persons into 3 groups, when the ordering of groups is not important can be done in \(\frac{1}{3!} * \frac{9!}{(3!)^3}\) ways. Answer is (A) or 280
_________________
Math writeups 1) Algebra101 2) Sequences 3) Set combinatorics 4) 3D geometry
My GMAT story
GMAT Club Premium Membership  big benefits and savings



Manager
Joined: 22 Aug 2008
Posts: 156

Re: 9 people and Combinatorics
[#permalink]
Show Tags
03 Oct 2010, 04:25
the number of ways to choose 9 people in 3 groups each having 3 people is
9C3 * 6C3 * 3C3 = 280
another way is = (3*3)!/((3!)^3)*3! = 280



Manager
Joined: 19 Apr 2011
Posts: 88

Re: 9 people and Combinatorics
[#permalink]
Show Tags
15 Jun 2011, 06:34
Hi Bunuel Can u pls explain why u r dividing by 3! we are not ordering here we are only choosing.



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8288
Location: Pune, India

Re: 9 people and Combinatorics
[#permalink]
Show Tags
15 Jun 2011, 20:06
toughmat wrote: Hi Bunuel Can u pls explain why u r dividing by 3! we are not ordering here we are only choosing. We divide by 3! because of exactly what you said: "we are not ordering here we are only choosing." When you say, "9C3 * 6C3 * 3C3," what you are doing is that you are choosing 3 people of 9 for group 1, 3 people out of the leftover 6 people for group 2 and the rest of the three people for group 3. You have inadvertently marked the 3 groups as distinct. But if we want to just divide them in 3 groups without any distinction of group 1, 2 or 3, we need to divide this by 3!.
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
GMAT selfstudy has never been more personalized or more fun. Try ORION Free!



Manager
Joined: 16 Feb 2011
Posts: 191
Schools: ABCD

Re: 9 people and Combinatorics
[#permalink]
Show Tags
18 Jun 2011, 22:14
How can we compute the # of ways in which 9 objects are divided into groups of 4,3 and 2 ? Can you please help ? Here's what I think: let 9 objects be AAAABBBCC Therefore, combinations = 9!/(4!*3!*2!) Correct ? [Order is important] If order is not important, since we have three groups, # of combinations = 9!/[(4!*3!*2!) * (3!)] Correct ?



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8288
Location: Pune, India

Re: 9 people and Combinatorics
[#permalink]
Show Tags
19 Jun 2011, 18:53
voodoochild wrote: How can we compute the # of ways in which 9 objects are divided into groups of 4,3 and 2 ? Can you please help ? Here's what I think: let 9 objects be AAAABBBCC Therefore, combinations = 9!/(4!*3!*2!) Correct ? [Order is important] If order is not important, since we have three groups, # of combinations = 9!/[(4!*3!*2!) * (3!)] Correct ? Actually, in this case the groups are distinct  a group of 4 people, another of 3 people and another of 2 people. A case in which Mr A is in the four person group is different from the one in which he is in 3 person group. So you will not divide by 3! at the end in second case.
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
GMAT selfstudy has never been more personalized or more fun. Try ORION Free!



Manager
Joined: 16 Feb 2011
Posts: 191
Schools: ABCD

Re: 9 people and Combinatorics
[#permalink]
Show Tags
19 Jun 2011, 19:51
VeritasPrepKarishma wrote: voodoochild wrote: How can we compute the # of ways in which 9 objects are divided into groups of 4,3 and 2 ? Can you please help ? Here's what I think: let 9 objects be AAAABBBCC Therefore, combinations = 9!/(4!*3!*2!) Correct ? [Order is important] If order is not important, since we have three groups, # of combinations = 9!/[(4!*3!*2!) * (3!)] Correct ? Actually, in this case the groups are distinct  a group of 4 people, another of 3 people and another of 2 people. A case in which Mr A is in the four person group is different from the one in which he is in 3 person group. So you will not divide by 3! at the end in second case. Thanks Karishma. So, are you saying that the order will not matter ? Essentially, # of combinations = 9!/(4!*3!*2!) irrespective of order?/ Thanks Voodoo



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8288
Location: Pune, India

Re: 9 people and Combinatorics
[#permalink]
Show Tags
21 Jun 2011, 02:29
voodoochild wrote: VeritasPrepKarishma wrote: voodoochild wrote: How can we compute the # of ways in which 9 objects are divided into groups of 4,3 and 2 ? Can you please help ? Here's what I think: let 9 objects be AAAABBBCC Therefore, combinations = 9!/(4!*3!*2!) Correct ? [Order is important] If order is not important, since we have three groups, # of combinations = 9!/[(4!*3!*2!) * (3!)] Correct ? Actually, in this case the groups are distinct  a group of 4 people, another of 3 people and another of 2 people. A case in which Mr A is in the four person group is different from the one in which he is in 3 person group. So you will not divide by 3! at the end in second case. Thanks Karishma. So, are you saying that the order will not matter ? Essentially, # of combinations = 9!/(4!*3!*2!) irrespective of order?/ Thanks Voodoo Yes, the groups are distinct so no of combinations is 9!/(4!*3!*2!)
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
GMAT selfstudy has never been more personalized or more fun. Try ORION Free!



Intern
Joined: 31 Oct 2011
Posts: 39
Concentration: General Management, Entrepreneurship
GPA: 3.4
WE: Accounting (Commercial Banking)

Re: 9 people and Combinatorics
[#permalink]
Show Tags
07 Oct 2012, 10:08
geturdream wrote: the number of ways to choose 9 people in 3 groups each having 3 people is
9C3 * 6C3 * 3C3 = 280
another way is = (3*3)!/((3!)^3)*3! = 280 Hi Geturdream, How comes A = 9C3 * 6C3 * 3C3 = 280 ? 9C3 = 9!/ (6!*3!) = 9*8*7 / (3*2) = 84 6C3 = 6! / (3!*3!) = 6*5*4 / (3*2) = 20 3C3 = 3!/ (3!*0!) = 1 => A = 84 * 20 = 1680 ?



Intern
Joined: 01 Jun 2011
Posts: 7

Re: In how many different ways can a group of 9 people be divide
[#permalink]
Show Tags
28 Oct 2013, 12:44
hemanthp wrote: In how many different ways can a group of 9 people be divided into 3 groups, with each group containing 3 people?
A. 280 B. 1,260 C. 1,680 D. 2,520 E. 3,360 I used a way found in another topic: How many ways can 1 person be put with the other 8 in groups of 3? 28 How many ways can following person be put with the remaining 5 in groups of 3? (subract first group total) : 10 How many ways can the final 3 be placed into a group of 3? (subtract last 3) : 1 28 * 10 * 1 = 280 answer is A



Intern
Joined: 07 Jul 2015
Posts: 2

Re: In how many different ways can a group of 9 people be divide
[#permalink]
Show Tags
25 Jul 2015, 09:11
gamelord wrote: geturdream wrote: the number of ways to choose 9 people in 3 groups each having 3 people is
9C3 * 6C3 * 3C3 = 280
another way is = (3*3)!/((3!)^3)*3! = 280 Hi Geturdream, How comes A = 9C3 * 6C3 * 3C3 = 280 ? 9C3 = 9!/ (6!*3!) = 9*8*7 / (3*2) = 84 6C3 = 6! / (3!*3!) = 6*5*4 / (3*2) = 20 3C3 = 3!/ (3!*0!) = 1 => A = 84 * 20 = 1680 ? 1680 needs to be divided by 3! because the order of the 3 groups do not matter for this problem. The order of the groups, i.e. G1G2G3 vs G2G3G1 (there are 4 more possible combinations) do not matter. \(\frac{9C3 * 6C3 * 3C3}{3!} = \frac{84*20*1}{3!} = \frac{1680}{3!} = 280\)



Intern
Joined: 18 Jul 2017
Posts: 1

Re: In how many different ways can a group of 9 people be divide
[#permalink]
Show Tags
19 Jul 2017, 01:16
Bunuel wrote: GENERAL RULE: 1. The number of ways in which mn different items can be divided equally into m groups, each containing n objects and the order of the groups is important is \(\frac{(mn)!}{(n!)^m}\) 2. The number of ways in which mn different items can be divided equally into m groups, each containing n objects and the order of the groups is NOT important is \(\frac{(mn)!}{(n!)^m * m!}\) I don't think that's right. For example, let's find out in how many ways 4 different items can be split into 2 groups of 2, when the order is NOT important, using the 2nd formula. \(\frac{(mn)!}{(n!)^m*m!}\) = \(\frac{(4)!}{(2!)^2*2!}\) = 3 This result can be disproved very easily graphically: 1. Draw a square 2. Draw an x through the square drawing 2 lines, corner to corner 3. Represent the 4 objects on each corner of the square 4. Number the lines that link all the square's corners together There are 6 lines which represent all 6 possible groups of 2 objects; every object is connected by a line to all other objects a single time, forming 6 possible groups of 2. The answer is 6. Another solution is 4C2 * 2C2 = 6 nCr = \(\frac{(n!)}{(r!(nr)!)}\) = \(\frac{(4!)}{(r!(42)!)}\) * \(\frac{(2!)}{(r!(22)!)}\) = 6 The groups have not been ordered yet and there is no need to divide by anything else; we proved graphically that 6 is the answer so dividing by 2! or anything else would lead to an incorrect result. If the order was important, there would be \(6!/(62)! = 30\) ways to organize 4 objects into 2 groups of 2.[/b] There would be 2 positions to fill with 6 possible groups. 6 possible groups in the 1st position and 5 possible groups in the 2nd position. nPr= \(\frac{n!}{(nr)!}\) = \(\frac{6!}{(62)!}\) = 6 * 5 = 30 The 1st formula actually represents the number of ways in which the order is NOT important\(\frac{(mn)!}{(n!)^m}\) = \(\frac{(4)!}{(2!)^2}\) = 6 THE ORIGINAL QUESTION:The order is not important in this question.nCr = \(\frac{(n!)}{(r!(nr)!)}\) = \(\frac{(9!)}{(3!(93)!)}\) * \(\frac{(6!)}{(3!(63)!)}\) * \(\frac{(3!)}{(3!(33)!)}\) = 1680 or, alternatively: \(\frac{(mn)!}{(n!)^m}\) = \(\frac{(9)!}{(3!)^3}\) = 1680 If the order mattered:nPr= \(\frac{n!}{(nr)!}\) = \(\frac{1680!}{(16803)!}\) = 1680 * 1679 * 1678 = 4733168160



Math Expert
Joined: 02 Sep 2009
Posts: 49303

In how many different ways can a group of 9 people be divide
[#permalink]
Show Tags
19 Jul 2017, 01:34
dontcountonme wrote: Bunuel wrote: GENERAL RULE: 1. The number of ways in which mn different items can be divided equally into m groups, each containing n objects and the order of the groups is important is \(\frac{(mn)!}{(n!)^m}\) 2. The number of ways in which mn different items can be divided equally into m groups, each containing n objects and the order of the groups is NOT important is \(\frac{(mn)!}{(n!)^m * m!}\) I don't think that's right. For example, let's find out in how many ways 4 different items can be split into 2 groups of 2, when the order is NOT important, using the 2nd formula. \(\frac{(mn)!}{(n!)^m*m!}\) = \(\frac{(4)!}{(2!)^2*2!}\) = 3 This result can be disproved very easily graphically: 1. Draw a square 2. Draw an x through the square drawing 2 lines, corner to corner 3. Represent the 4 objects on each corner of the square 4. Number the lines that link all the square's corners together There are 6 lines which represent all 6 possible groups of 2 objects; every object is connected by a line to all other objects a single time, forming 6 possible groups of 2. The answer is 6. No you are not right. There are two formulas. The first one for which the order of the groups IS important and the second one for which the order of the groups is NOT important. Case 1  the order of the groups IS important: \(\frac{(mn)!}{(n!)^m}\)2*2 = 4 different items can be divided equally into 2 groups and the order of the groups IS important \(\frac{4!}{(2!)^2}=6\). GROUP 1  GROUP 2 {AB}  {CD} {CD}  {AB} {AC}  {BD} {BD}  {AC} {AD}  {BC} {BC}  {AD} Case 2  the order of the groups is NOT important: \(\frac{(mn)!}{(n!)^m * m!}\)2*2 = 4 different items can be divided equally into 2 groups and the order of the groups is NOT important \(\frac{4!}{(2!)^2*2}=3\). {AB}  {CD} {AC}  {BD} {AD}  {BC} P.S. The correct answer is A, not C. Please refer to several different solutions presented above.
_________________
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



Manager
Joined: 27 Jan 2016
Posts: 144

Re: In how many different ways can a group of 9 people be divide
[#permalink]
Show Tags
03 Aug 2017, 05:03
VeritasPrepKarishma wrote: toughmat wrote: Hi Bunuel Can u pls explain why u r dividing by 3! we are not ordering here we are only choosing. We divide by 3! because of exactly what you said: "we are not ordering here we are only choosing." When you say, "9C3 * 6C3 * 3C3," what you are doing is that you are choosing 3 people of 9 for group 1, 3 people out of the leftover 6 people for group 2 and the rest of the three people for group 3. You have inadvertently marked the 3 groups as distinct. But if we want to just divide them in 3 groups without any distinction of group 1, 2 or 3, we need to divide this by 3!. Can we apply the same logic to the below question. A menu consisting of 2 types of breakfasts, 3 types of lunch and 4 types of dinner has to be selected from 10 types of breakfast, 11 types of lunch and 12 types of dinner. Will the solution be 10c*11c3*12c4/3! ?



Manager
Joined: 27 Jan 2016
Posts: 144

Re: In how many different ways can a group of 9 people be divide
[#permalink]
Show Tags
03 Aug 2017, 05:05
Bunuel wrote: dontcountonme wrote: Bunuel wrote: GENERAL RULE: 1. The number of ways in which mn different items can be divided equally into m groups, each containing n objects and the order of the groups is important is \(\frac{(mn)!}{(n!)^m}\) 2. The number of ways in which mn different items can be divided equally into m groups, each containing n objects and the order of the groups is NOT important is \(\frac{(mn)!}{(n!)^m * m!}\) I don't think that's right. For example, let's find out in how many ways 4 different items can be split into 2 groups of 2, when the order is NOT important, using the 2nd formula. \(\frac{(mn)!}{(n!)^m*m!}\) = \(\frac{(4)!}{(2!)^2*2!}\) = 3 This result can be disproved very easily graphically: 1. Draw a square 2. Draw an x through the square drawing 2 lines, corner to corner 3. Represent the 4 objects on each corner of the square 4. Number the lines that link all the square's corners together There are 6 lines which represent all 6 possible groups of 2 objects; every object is connected by a line to all other objects a single time, forming 6 possible groups of 2. The answer is 6. No you are not right. There are two formulas. The first one for which the order of the groups IS important and the second one for which the order of the groups is NOT important. Case 1  the order of the groups IS important: \(\frac{(mn)!}{(n!)^m}\)2*2 = 4 different items can be divided equally into 2 groups and the order of the groups IS important \(\frac{4!}{(2!)^2}=6\). GROUP 1  GROUP 2 {AB}  {CD} {CD}  {AB} {AC}  {BD} {BD}  {AC} {AD}  {BC} {BC}  {AD} Case 2  the order of the groups is NOT important: \(\frac{(mn)!}{(n!)^m * m!}\)2*2 = 4 different items can be divided equally into 2 groups and the order of the groups is NOT important \(\frac{4!}{(2!)^2*2}=3\). {AB}  {CD} {AC}  {BD} {AD}  {BC} P.S. The correct answer is A, not C. Please refer to several different solutions presented above. Can we apply the same logic to the below question. A menu consisting of 2 types of breakfasts, 3 types of lunch and 4 types of dinner has to be selected from 10 types of breakfast, 11 types of lunch and 12 types of dinner. Will the solution be 10c*11c3*12c4/3! ?



Senior Manager
Joined: 08 Jun 2013
Posts: 343
Location: India
GPA: 3.82
WE: Engineering (Other)

Re: In how many different ways can a group of 9 people be divide
[#permalink]
Show Tags
22 Sep 2018, 18:59
We can start with this combination problem, like most of them, by looking at one part of it. How many ways are there of forming the first group or pile? That’s choosing 3 people from 9. Applying the combination formula, with n = 9 and k = 3, \(9C3\) = 84 To form the next group, we compute \(6C3\) = 20 There are three people from the original group remaining, and only one group of three left, so there is only one way to finish. That means there are 84 × 20 × 1 = 1,680 ways to create the three groups. However, that number, 1,680, involves doublecounting some groups: it counts ABC DEF GHI and DEF ABC GHI as different ways to do the grouping, but for our purposes those ways are the same, since the three groups that we are placing people into are indistinguishable. For each way, the duplication will be the number of ways of shuffling the three groups around, which is 3! = 6. The number of unique ways to form the groups is therefore . The Correct Answer is (A).
_________________
It seems Kudos button not working correctly with all my posts...
Please check if it is working with this post......
is it?....
Anyways...Thanks for trying




Re: In how many different ways can a group of 9 people be divide &nbs
[#permalink]
22 Sep 2018, 18:59






