It is currently 22 Nov 2017, 02:48

### 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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Nine dogs are split into 3 groups to pull one of three

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

### Hide Tags

Intern
Joined: 23 Aug 2009
Posts: 16

Kudos [?]: 21 [0], given: 8

Nine dogs are split into 3 groups to pull one of three [#permalink]

### Show Tags

03 Jan 2010, 06:03
21
This post was
BOOKMARKED
00:00

Difficulty:

(N/A)

Question Stats:

57% (00:54) correct 43% (00:22) wrong based on 33 sessions

### HideShow timer Statistics

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?
_________________

Peace

Kudos [?]: 21 [0], given: 8

Math Expert
Joined: 02 Sep 2009
Posts: 42302

Kudos [?]: 133013 [5], given: 12402

Re: Probability [#permalink]

### Show Tags

03 Jan 2010, 08:49
5
This post received
KUDOS
Expert's post
20
This post was
BOOKMARKED
ro86 wrote:
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.
_________________

Kudos [?]: 133013 [5], given: 12402

Intern
Joined: 23 Aug 2009
Posts: 16

Kudos [?]: 21 [0], given: 8

Re: Probability [#permalink]

### Show Tags

03 Jan 2010, 09:06
Thanks.
So if there are any tasks assigned to the groups then we will have to consider the first case.
_________________

Peace

Kudos [?]: 21 [0], given: 8

Math Expert
Joined: 02 Sep 2009
Posts: 42302

Kudos [?]: 133013 [1], given: 12402

Re: Probability [#permalink]

### Show Tags

04 Jan 2010, 00:10
1
This post received
KUDOS
Expert's post
ro86 wrote:
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!
_________________

Kudos [?]: 133013 [1], given: 12402

Manager
Status: Last few days....Have pressed the throttle
Joined: 20 Jun 2010
Posts: 67

Kudos [?]: 62 [0], given: 27

WE 1: 6 years - Consulting
Re: Probability [#permalink]

### Show Tags

19 Aug 2010, 19:32
Bunuel wrote:
ro86 wrote:
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

Kudos [?]: 62 [0], given: 27

Math Expert
Joined: 02 Sep 2009
Posts: 42302

Kudos [?]: 133013 [1], given: 12402

Re: Probability [#permalink]

### Show Tags

20 Aug 2010, 08:28
1
This post received
KUDOS
Expert's post
oldstudent wrote:
Bunuel wrote:
ro86 wrote:
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

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.
_________________

Kudos [?]: 133013 [1], given: 12402

Manager
Joined: 02 Oct 2010
Posts: 145

Kudos [?]: 50 [0], given: 29

Re: Probability [#permalink]

### Show Tags

07 Jan 2011, 23:03
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...

Kudos [?]: 50 [0], given: 29

Math Expert
Joined: 02 Sep 2009
Posts: 42302

Kudos [?]: 133013 [0], given: 12402

Re: Probability [#permalink]

### Show Tags

08 Jan 2011, 03:31
Expert's post
3
This post was
BOOKMARKED
jullysabat wrote:
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...

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.
_________________

Kudos [?]: 133013 [0], given: 12402

Current Student
Status: Everyone is a leader. Just stop listening to others.
Joined: 22 Mar 2013
Posts: 956

Kudos [?]: 1899 [0], given: 229

Location: India
GPA: 3.51
WE: Information Technology (Computer Software)
Re: Nine dogs are split into 3 groups to pull one of three [#permalink]

### Show Tags

24 Jul 2013, 13:24
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.
_________________

Piyush K
-----------------------
Our greatest weakness lies in giving up. The most certain way to succeed is to try just one more time. ― Thomas A. Edison
Don't forget to press--> Kudos
My Articles: 1. WOULD: when to use? | 2. All GMATPrep RCs (New)
Tip: Before exam a week earlier don't forget to exhaust all gmatprep problems specially for "sentence correction".

Kudos [?]: 1899 [0], given: 229

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7745

Kudos [?]: 17843 [1], given: 235

Location: Pune, India
Re: Nine dogs are split into 3 groups to pull one of three [#permalink]

### Show Tags

24 Jul 2013, 23:08
1
This post received
KUDOS
Expert's post
PiyushK wrote:
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.

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
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for \$199

Veritas Prep Reviews

Kudos [?]: 17843 [1], given: 235

Non-Human User
Joined: 09 Sep 2013
Posts: 15564

Kudos [?]: 283 [0], given: 0

Re: Nine dogs are split into 3 groups to pull one of three [#permalink]

### Show Tags

23 Aug 2014, 04:00
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.
_________________

Kudos [?]: 283 [0], given: 0

Non-Human User
Joined: 09 Sep 2013
Posts: 15564

Kudos [?]: 283 [1], given: 0

Re: Nine dogs are split into 3 groups to pull one of three [#permalink]

### Show Tags

11 Oct 2015, 04:59
1
This post received
KUDOS
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.
_________________

Kudos [?]: 283 [1], given: 0

Intern
Joined: 16 Nov 2015
Posts: 17

Kudos [?]: 5 [0], given: 64

Nine dogs are split into 3 groups to pull one of three [#permalink]

### Show Tags

26 Jan 2016, 07:38
Bunuel wrote:
ro86 wrote:
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!

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

Kudos [?]: 5 [0], given: 64

Non-Human User
Joined: 09 Sep 2013
Posts: 15564

Kudos [?]: 283 [0], given: 0

Re: Nine dogs are split into 3 groups to pull one of three [#permalink]

### Show Tags

19 Feb 2017, 07:57
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.
_________________

Kudos [?]: 283 [0], given: 0

Re: Nine dogs are split into 3 groups to pull one of three   [#permalink] 19 Feb 2017, 07:57
Display posts from previous: Sort by

# Nine dogs are split into 3 groups to pull one of three

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

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.