Last visit was: 19 Nov 2025, 16:26 It is currently 19 Nov 2025, 16:26
Home
GMAT
Messages
Tests
Schools
Events
Rewards
Chat
My Profile
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.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
User avatar
ro86
User avatar
Joined: 23 Aug 2009
Last visit: 11 May 2010
Posts: 8
Own Kudos:
70
 [57]
Given Kudos: 8
Posts: 8
Kudos: 70
 [57]
Nine dogs are split into 3 groups to pull one of three 03 Jan 2010, 06:03
2
Kudos
Add Kudos
53
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
N
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

(N/A)

Question Stats:

62% (00:52) correct 38% (01:30) wrong based on 103 sessions

History

Date
Time
Result
Not Attempted Yet
1. Nine dogs are split into 3 groups to pull one of three sleds in a race. How many different assignments of dogs to sleds are possible?

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

Are both the questions the same?
Most Helpful Reply
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,390
Own Kudos:
778,368
 [66]
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,390
Kudos: 778,368
 [66]
Nine dogs are split into 3 groups to pull one of three 03 Jan 2010, 08:49
9
Kudos
Add Kudos
55
Bookmarks
Bookmark this Post
ro86
1. Nine dogs are split into 3 groups to pull one of three sleds in a race. How many different assignments of dogs to sleds are possible?

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

Are both the questions the same?
Show more

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 \(\frac{(mn)!}{(n!)^m*m!}\).


In the first case: I think the order is important, as we'll have group #1, #2 and #3 assigned to specific task. So we should use first formula, mn=9, m=3 groups n=3 objects (dogs):
\(\frac{(mn)!}{(n!)^m}=\frac{9!}{(3!)^3}=1680\).

This can be done in another way as well: \(9C3*6C3*3C3=1680\), (9C3 # of ways of choosing 3 from 9, 6C3 # of ways of choosing 3 from 6, 3C3 # of ways of choosing 3 from 3).


In the second case: I think the order is NOT important, as we won't have group #1, #2 and #3. So we should use second formula, again \(mn=9\), \(m=3\) 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.
General Discussion
User avatar
ro86
User avatar
Joined: 23 Aug 2009
Last visit: 11 May 2010
Posts: 8
Own Kudos:
70
 [1]
Given Kudos: 8
Posts: 8
Kudos: 70
 [1]
Nine dogs are split into 3 groups to pull one of three 03 Jan 2010, 09:06
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Thanks.
So if there are any tasks assigned to the groups then we will have to consider the first case.
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,390
Own Kudos:
778,368
 [1]
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,390
Kudos: 778,368
 [1]
Nine dogs are split into 3 groups to pull one of three 04 Jan 2010, 00:10
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
ro86
Thanks.
So if there are any tasks assigned to the groups then we will have to consider the first case.
Show more

Yes. Consider the following example: there are three teams and three tournaments. In how many different ways can we assign the teams to tournaments?

Answer: 3!
User avatar
oldstudent
User avatar
Joined: 20 Jun 2010
Last visit: 31 Jul 2015
Posts: 39
Own Kudos:
217
 [1]
Given Kudos: 27
Status:Last few days....Have pressed the throttle
Concentration: Entrepreneurship,Strategy, Marketing
WE 1: 6 years - Consulting
Posts: 39
Kudos: 217
 [1]
Nine dogs are split into 3 groups to pull one of three 19 Aug 2010, 19:32
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
ro86
Thanks.
So if there are any tasks assigned to the groups then we will have to consider the first case.

Yes. Consider the following example: there are three teams and three tournaments. In how many different ways can we assign the teams to tournaments?

Answer: 3!
Show more

Hi Bunuel,

Please explain how did you get 3! . I am not able to understand how to get this by the above (subgroup) formula.
Thanks
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,390
Own Kudos:
778,368
 [2]
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,390
Kudos: 778,368
 [2]
Nine dogs are split into 3 groups to pull one of three 20 Aug 2010, 08:28
1
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
oldstudent
Bunuel
ro86
Thanks.
So if there are any tasks assigned to the groups then we will have to consider the first case.

Yes. Consider the following example: there are three teams and three tournaments. In how many different ways can we assign the teams to tournaments?

Answer: 3!

Hi Bunuel,

Please explain how did you get 3! . I am not able to understand how to get this by the above (subgroup) formula.
Thanks
Show more

Tournaments: A, B, C.
Teams: 1, 2, 3.

# of assignments of teams to tournaments is 3!.
ABC
123
132
213
231
312
321

Please check the following link: combinations-problems-95344.html?hilit=factorial%20teams#p734396

Hope it helps.
User avatar
jullysabat
User avatar
Joined: 02 Oct 2010
Last visit: 08 May 2012
Posts: 67
Own Kudos:
52
Given Kudos: 29
Posts: 67
Kudos: 52
Nine dogs are split into 3 groups to pull one of three 07 Jan 2011, 23:03
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hello Bunnel,

In one of the Qs you had explained like this----
In how many different ways can a group of 8 people be divided into 4 teams of 2 people each?
90
105
168
420
2520

AS:
For the first person we can pick a pair in 7 ways;
For the second one in 5 ways (as two are already chosen);
For the third one in 3 ways (as 4 people are already chosen);
For the fourth one there is only one left.
So we have 7*5*3*1=105

Similarly, for this solution we can think as

For the first person we can pick three people in 8 ways(As there are 9 people);
For the second one in 5 ways (as three are already chosen);
For the third one in 2 ways (as 6 people are already chosen);
So we have 8*5*2=80 ways...

But the answer here is 280... form combination formula..
I find both the patterns same so I did it in that way...But The answer I get is 80 not 280...
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,390
Own Kudos:
778,368
 [4]
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,390
Kudos: 778,368
 [4]
Nine dogs are split into 3 groups to pull one of three 08 Jan 2011, 03:31
Kudos
Add Kudos
4
Bookmarks
Bookmark this Post
jullysabat
Hello Bunnel,

In one of the Qs you had explained like this----
In how many different ways can a group of 8 people be divided into 4 teams of 2 people each?
90
105
168
420
2520

AS:
For the first person we can pick a pair in 7 ways;
For the second one in 5 ways (as two are already chosen);
For the third one in 3 ways (as 4 people are already chosen);
For the fourth one there is only one left.
So we have 7*5*3*1=105

Similarly, for this solution we can think as

For the first person we can pick three people in 8 ways(As there are 9 people);
For the second one in 5 ways (as three are already chosen);
For the third one in 2 ways (as 6 people are already chosen);
So we have 8*5*2=80 ways...

But the answer here is 280... form combination formula..
I find both the patterns same so I did it in that way...But The answer I get is 80 not 280...
Show more

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

There should be 3 people in 3 groups:

For the first person we can pick TWO partners in \(C^2_8=28\) ways;
For the second one we can pick TWO partners in \(C^2_5=10\) ways (as 3 people are already chosen);
For the third one we can pick TWO partners in \(C^2_2=1\) ways (as 6 people are already chosen);;

So we have 28*10=280.

Hope it's clear.
User avatar
PiyushK
User avatar
Joined: 22 Mar 2013
Last visit: 31 Aug 2025
Posts: 598
Own Kudos:
4,978
Given Kudos: 235
Status:Everyone is a leader. Just stop listening to others.
Location: India
GPA: 3.51
WE:Information Technology (Computer Software)
Products:
My Reviews
Posts: 598
Kudos: 4,978
Nine dogs are split into 3 groups to pull one of three 24 Jul 2013, 13:24
Kudos
Add Kudos
Bookmarks
Bookmark this Post
1. Nine dogs are split into 3 groups to pull one of three sleds in a race. How many different assignments of dogs to sleds are possible?

This question is not stating that equal assignment is expected in each group, thus application of following formula looks appropriate to distribute n identical objects among r number of receivers such that at least one object goes to everyone.

n-1Cr-1 = 8C2 = 28.
User avatar
KarishmaB
User avatar
Joined: 16 Oct 2010
Last visit: 19 Nov 2025
Posts: 16,267
Own Kudos:
77,001
 [2]
Given Kudos: 482
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,267
Kudos: 77,001
 [2]
Nine dogs are split into 3 groups to pull one of three 24 Jul 2013, 23:08
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
PiyushK
1. Nine dogs are split into 3 groups to pull one of three sleds in a race. How many different assignments of dogs to sleds are possible?

This question is not stating that equal assignment is expected in each group, thus application of following formula looks appropriate to distribute n identical objects among r number of receivers such that at least one object goes to everyone.

n-1Cr-1 = 8C2 = 28.
Show more

This formula is used when you have n identical objects to be distributed in r distinct groups. Here, it is not given that the dogs are identical. Just like with 9 people, you don't assume that they are identical, similarly, with 9 dogs you cannot assume so. Things e.g. fruits (9 apples) may be considered identical but the word identical will be mentioned for clarity. Also, it is kind of implied that you need 3 dogs per sled so 9 dogs need to be split into 3 groups of 3 dogs each to pull the 3 sleds. I agree that it is not given clearly that each sled needs 3 dogs and an actual GMAT question will do justice.

The way to go about this question is 9C3*6C3*3C3
User avatar
leve
User avatar
Joined: 16 Nov 2015
Last visit: 26 May 2016
Posts: 13
Own Kudos:
16
Given Kudos: 64
Posts: 13
Kudos: 16
Nine dogs are split into 3 groups to pull one of three 26 Jan 2016, 07:38
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
ro86
Thanks.
So if there are any tasks assigned to the groups then we will have to consider the first case.

Yes. Consider the following example: there are three teams and three tournaments. In how many different ways can we assign the teams to tournaments?

Answer: 3!
Show more

Bunuel, in this question dont you have to mention that each team will be placed in one tournament only? Other wise we can think that each tournement can take more than one team. (For example Team A and B to Tournament 1, Team C to Tournament 2 and no teams to Tournament 3 )

This way the answer should be 3^3 = 27 right? Please correct me if i am wrong.

Thanx
User avatar
gaurav2m
User avatar
Joined: 15 Apr 2018
Last visit: 22 Mar 2021
Posts: 62
Own Kudos:
39
Given Kudos: 382
Location: India
Schools: NUS '20
Schools: NUS '20
Posts: 62
Kudos: 39
Nine dogs are split into 3 groups to pull one of three 07 Mar 2020, 06:05
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
ro86
1. Nine dogs are split into 3 groups to pull one of three sleds in a race. How many different assignments of dogs to sleds are possible?

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

Are both the questions the same?

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 \(\frac{(mn)!}{(n!)^m*m!}\).



In the first case: I think the order is important, as we'll have group #1, #2 and #3 assigned to specific task. So we should use first formula, mn=9, m=3 groups n=3 objects (dogs):
\(\frac{(mn)!}{(n!)^m}=\frac{9!}{(3!)^3}=1680\).

This can be done in another way as well: \(9C3*6C3*3C3=1680\), (9C3 # of ways of choosing 3 from 9, 6C3 # of ways of choosing 3 from 6, 3C3 # of ways of choosing 3 from 3).


In the second case: I think the order is NOT important, as we won't have group #1, #2 and #3. So we should use second formula, again \(mn=9\), \(m=3\) 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.
Show more


hi Bunuel,

how are you dividing groups in the first case with 3 members each?
avatar
jayarora
avatar
Current Student
Joined: 24 Oct 2016
Last visit: 26 Apr 2025
Posts: 163
Own Kudos:
237
Given Kudos: 116
Location: India
Concentration: Technology, Strategy
Schools: Harvard '24 (A)
GMAT 1: 710 Q49 V38
GMAT 2: 760 Q50 V44 (Online)
GPA: 3.61
Products:
My Reviews
Schools: Harvard '24 (A)
GMAT 2: 760 Q50 V44 (Online)
Posts: 163
Kudos: 237
Nine dogs are split into 3 groups to pull one of three 08 Mar 2020, 04:52
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
ro86
1. Nine dogs are split into 3 groups to pull one of three sleds in a race. How many different assignments of dogs to sleds are possible?

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

Are both the questions the same?

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 \(\frac{(mn)!}{(n!)^m*m!}\).


In the first case: I think the order is important, as we'll have group #1, #2 and #3 assigned to specific task. So we should use first formula, mn=9, m=3 groups n=3 objects (dogs):
\(\frac{(mn)!}{(n!)^m}=\frac{9!}{(3!)^3}=1680\).

This can be done in another way as well: \(9C3*6C3*3C3=1680\), (9C3 # of ways of choosing 3 from 9, 6C3 # of ways of choosing 3 from 6, 3C3 # of ways of choosing 3 from 3).


In the second case: I think the order is NOT important, as we won't have group #1, #2 and #3. So we should use second formula, again \(mn=9\), \(m=3\) 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.
Show more


Hi

Why don't we multiple by 3! again in the first case 9C3∗6C3∗3C3=1680 to shuffle the 3 groups and assign them to the sledges?

Thanks
User avatar
gaurav2m
User avatar
Joined: 15 Apr 2018
Last visit: 22 Mar 2021
Posts: 62
Own Kudos:
39
Given Kudos: 382
Location: India
Schools: NUS '20
Schools: NUS '20
Posts: 62
Kudos: 39
Nine dogs are split into 3 groups to pull one of three 09 Mar 2020, 05:55
Kudos
Add Kudos
Bookmarks
Bookmark this Post
jayarora
Bunuel
ro86
1. Nine dogs are split into 3 groups to pull one of three sleds in a race. How many different assignments of dogs to sleds are possible?

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

Are both the questions the same?

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 \(\frac{(mn)!}{(n!)^m*m!}\).



In the first case: I think the order is important, as we'll have group #1, #2 and #3 assigned to specific task. So we should use first formula, mn=9, m=3 groups n=3 objects (dogs):
\(\frac{(mn)!}{(n!)^m}=\frac{9!}{(3!)^3}=1680\).

This can be done in another way as well: \(9C3*6C3*3C3=1680\), (9C3 # of ways of choosing 3 from 9, 6C3 # of ways of choosing 3 from 6, 3C3 # of ways of choosing 3 from 3).


In the second case: I think the order is NOT important, as we won't have group #1, #2 and #3. So we should use second formula, again \(mn=9\), \(m=3\) 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.


Hi

Why don't we multiple by 3! again in the first case 9C3∗6C3∗3C3=1680 to shuffle the 3 groups and assign them to the sledges?

Thanks
Show more


Hi jay,

In the first case order matters because we are assigning some tasks in 'n' group. Therefore, in this case, 9C3∗6C3∗3C3 all the scenarios are covered such as (ABC, DEF, GHI), (DEF, GHI ,ABC) etc. as arrangement is already done then there is no need to arrange the sequence again.

But yes, there are some scenarios in which order doesn't matter like in the second scenario. then you have to divide it by 3! to remove the duplicate sequence.
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 38,589
Own Kudos:
1,079
Posts: 38,589
Kudos: 1,079
Nine dogs are split into 3 groups to pull one of three 14 Nov 2024, 05:26
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Moderators:
Bunuel
Math Expert
105390 posts
Krunaal
Tuck School Moderator
805 posts