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

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Nine dogs are split into 3 groups to pull one of three [#permalink] New post 03 Jan 2010, 05:03
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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?

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Re: Probability [#permalink] New post 03 Jan 2010, 07:49
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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.

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Re: Probability [#permalink] New post 03 Jan 2010, 23:10
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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!

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Re: Probability [#permalink] New post 20 Aug 2010, 07:28
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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.

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Re: Nine dogs are split into 3 groups to pull one of three [#permalink] New post 24 Jul 2013, 22:08
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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

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Re: Probability [#permalink] New post 03 Jan 2010, 08:06
Thanks.
So if there are any tasks assigned to the groups then we will have to consider the first case.

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Re: Probability [#permalink] New post 19 Aug 2010, 18:32
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
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Re: Probability [#permalink] New post 07 Jan 2011, 22:03
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...
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Re: Probability [#permalink] New post 08 Jan 2011, 02:31
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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.

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RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

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.


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Re: Nine dogs are split into 3 groups to pull one of three [#permalink] New post 24 Jul 2013, 12:24
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.

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Re: Nine dogs are split into 3 groups to pull one of three   [#permalink] 24 Jul 2013, 12:24
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