GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 18 Sep 2019, 06:51

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

In how many different ways can a group of 9 people be divide

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Find Similar Topics 
Manager
Manager
avatar
Status: Keep fighting!
Affiliations: IIT Madras
Joined: 31 Jul 2010
Posts: 182
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

New post 26 Sep 2010, 09:47
12
51
00:00
A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

51% (01:32) correct 49% (01:51) wrong based on 449 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
Most Helpful Expert Reply
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58093
Re: 9 people and Combinatorics  [#permalink]

Show Tags

New post 26 Sep 2010, 09:56
9
33
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.
_________________
General Discussion
Retired Moderator
User avatar
Joined: 02 Sep 2010
Posts: 741
Location: London
GMAT ToolKit User Reviews Badge
Re: 9 people and Combinatorics  [#permalink]

Show Tags

New post 02 Oct 2010, 00:41
2
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
_________________
Manager
Manager
avatar
Joined: 22 Aug 2008
Posts: 113
Re: 9 people and Combinatorics  [#permalink]

Show Tags

New post 03 Oct 2010, 04:25
3
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
Manager
avatar
Joined: 19 Apr 2011
Posts: 81
Re: 9 people and Combinatorics  [#permalink]

Show Tags

New post 15 Jun 2011, 06:34
1
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
User avatar
D
Joined: 16 Oct 2010
Posts: 9637
Location: Pune, India
Re: 9 people and Combinatorics  [#permalink]

Show Tags

New post 15 Jun 2011, 20:06
8
3
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 >
Manager
Manager
avatar
Joined: 16 Feb 2011
Posts: 165
Schools: ABCD
Re: 9 people and Combinatorics  [#permalink]

Show Tags

New post 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
User avatar
D
Joined: 16 Oct 2010
Posts: 9637
Location: Pune, India
Re: 9 people and Combinatorics  [#permalink]

Show Tags

New post 19 Jun 2011, 18:53
1
1
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 >
Manager
Manager
avatar
Joined: 16 Feb 2011
Posts: 165
Schools: ABCD
Re: 9 people and Combinatorics  [#permalink]

Show Tags

New post 19 Jun 2011, 19:51
1
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
User avatar
D
Joined: 16 Oct 2010
Posts: 9637
Location: Pune, India
Re: 9 people and Combinatorics  [#permalink]

Show Tags

New post 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 >
Intern
Intern
avatar
Joined: 31 Oct 2011
Posts: 32
Concentration: General Management, Entrepreneurship
GMAT 1: 710 Q50 V35
GPA: 3.4
WE: Accounting (Commercial Banking)
Re: 9 people and Combinatorics  [#permalink]

Show Tags

New post 07 Oct 2012, 10:08
2
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
Intern
avatar
B
Joined: 01 Jun 2011
Posts: 15
Re: In how many different ways can a group of 9 people be divide  [#permalink]

Show Tags

New post 28 Oct 2013, 12:44
4
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
Intern
avatar
Joined: 07 Jul 2015
Posts: 2
Re: In how many different ways can a group of 9 people be divide  [#permalink]

Show Tags

New post 25 Jul 2015, 09:11
3
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
Intern
avatar
Joined: 18 Jul 2017
Posts: 1
Re: In how many different ways can a group of 9 people be divide  [#permalink]

Show Tags

New post 19 Jul 2017, 01:16
1
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 = \(\frac{(n!)}{(r!(n-r)!)}\) = \(\frac{(4!)}{(r!(4-2)!)}\) * \(\frac{(2!)}{(r!(2-2)!)}\) = 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 \(6!/(6-2)! = 30\) 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= \(\frac{n!}{(n-r)!}\) = \(\frac{6!}{(6-2)!}\) = 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 = \(\frac{(n!)}{(r!(n-r)!)}\) = \(\frac{(9!)}{(3!(9-3)!)}\) * \(\frac{(6!)}{(3!(6-3)!)}\) * \(\frac{(3!)}{(3!(3-3)!)}\) = 1680


or, alternatively:


\(\frac{(mn)!}{(n!)^m}\) = \(\frac{(9)!}{(3!)^3}\) = 1680

If the order mattered:

nPr= \(\frac{n!}{(n-r)!}\) = \(\frac{1680!}{(1680-3)!}\) = 1680 * 1679 * 1678 = 4733168160
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58093
In how many different ways can a group of 9 people be divide  [#permalink]

Show Tags

New post 19 Jul 2017, 01:34
2
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.
_________________
Manager
Manager
avatar
G
Joined: 27 Jan 2016
Posts: 125
Schools: ISB '18
GMAT 1: 700 Q50 V34
Re: In how many different ways can a group of 9 people be divide  [#permalink]

Show Tags

New post 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
Manager
avatar
G
Joined: 27 Jan 2016
Posts: 125
Schools: ISB '18
GMAT 1: 700 Q50 V34
Re: In how many different ways can a group of 9 people be divide  [#permalink]

Show Tags

New post 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! ?
Director
Director
User avatar
D
Joined: 08 Jun 2013
Posts: 548
Location: France
Schools: INSEAD Jan '19
GMAT 1: 200 Q1 V1
GPA: 3.82
WE: Consulting (Other)
GMAT ToolKit User
Re: In how many different ways can a group of 9 people be divide  [#permalink]

Show Tags

New post 22 Sep 2018, 18:59
1
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 double-counting 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).
_________________
Everything will fall into place…

There is perfect timing for
everything and everyone.
Never doubt, But Work on
improving yourself,
Keep the faith and
Stay ready. When it’s
finally your turn,
It will all make sense.
Target Test Prep Representative
User avatar
D
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 7735
Location: United States (CA)
Re: In how many different ways can a group of 9 people be divide  [#permalink]

Show Tags

New post 27 May 2019, 05:31
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


The first group of 3 can be chosen in 9C3 = (9 x 8 x 7)/(3 x 2) = 3 x 4 x 7 = 84 ways.

The second group of 3 can be chosen in 6C3 = (6 x 5 x 4)/(3 x 2) = 5 x 4 = 20 ways.

The third group of 3 can be chosen in 3C3 = 1 way.

Therefore, the 3 groups can be chosen 84 x 20 x 1 = 1680 ways. However, since the order of the 3 groups doesn’t matter, we have to divide 1680 by 3!. Hence, the number of ways 9 people can be divided into 3 groups is 1680/3! = 1680/6 = 280.

Answer: A
_________________

Scott Woodbury-Stewart

Founder and CEO

Scott@TargetTestPrep.com
TTP - Target Test Prep Logo
122 Reviews

5-star rated online GMAT quant
self study course

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button.

SVP
SVP
User avatar
P
Joined: 03 Jun 2019
Posts: 1512
Location: India
Premium Member Reviews Badge CAT Tests
In how many different ways can a group of 9 people be divide  [#permalink]

Show Tags

New post 13 Sep 2019, 12:58
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


Asked: In how many different ways can a group of 9 people be divided into 3 groups, with each group containing 3 people?

Number of ways = (9C3 * 6C3 * 3C3)/3! = (84 * 20 * 1)/6 = 1680/6 = 280

IMO A
_________________
"Success is not final; failure is not fatal: It is the courage to continue that counts."

Please provide kudos if you like my post. Kudos encourage active discussions.

My GMAT Resources: -

Efficient Learning
All you need to know about GMAT quant

Tele: +91-11-40396815
Mobile : +91-9910661622
E-mail : kinshook.chaturvedi@gmail.com
GMAT Club Bot
In how many different ways can a group of 9 people be divide   [#permalink] 13 Sep 2019, 12:58
Display posts from previous: Sort by

In how many different ways can a group of 9 people be divide

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  





Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne