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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. _________________

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 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);;

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

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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".

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

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24 Jul 2013, 23:08

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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 _________________

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

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23 Aug 2014, 04:00

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

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11 Oct 2015, 04:59

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

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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

gmatclubot

Nine dogs are split into 3 groups to pull one of three
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26 Jan 2016, 07:38

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