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Nine dogs are split into 3 groups to pull one of three
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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
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03 Jan 2010, 07:49
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
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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
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03 Jan 2010, 23:10
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
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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
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20 Aug 2010, 07:28
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!. ABC123 132 213 231 312 321 Please check the following link: combinationsproblems95344.html?hilit=factorial%20teams#p734396Hope it helps.
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Re: Probability
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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
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08 Jan 2011, 02:31
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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Re: Nine dogs are split into 3 groups to pull one of three
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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.
n1Cr1 = 8C2 = 28.



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Re: Nine dogs are split into 3 groups to pull one of three
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24 Jul 2013, 22:08
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.
n1Cr1 = 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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Nine dogs are split into 3 groups to pull one of three
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26 Jan 2016, 06:38
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



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Nine dogs are split into 3 groups to pull one of three
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07 Mar 2020, 05:05
Bunuel wrote: 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. hi Bunuel, how are you dividing groups in the first case with 3 members each?



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Re: Nine dogs are split into 3 groups to pull one of three
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08 Mar 2020, 03:52
Bunuel wrote: 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. 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
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Re: Nine dogs are split into 3 groups to pull one of three
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09 Mar 2020, 04:55
jayarora wrote: Bunuel wrote: 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. 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 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.




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