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Bunuel
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In a family with 3 children, the parents have agreed to bring the chil 04 Dec 2019, 01:08
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85% (hard)

Question Stats:

45% (02:17) correct 55% (02:04) wrong based on 227 sessions

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In a family with 3 children, the parents have agreed to bring the children to the pet store and allow each child to choose a pet. This pet store sells only dogs, cats, and monkeys. If each child chooses exactly one animal, and if more than one child can choose the same kind of animal, how many different arrangements of animals could the family leave with?

A. 6
B. 8
C. 9
D. 10
E. 12


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In a family with 3 children, the parents have agreed to bring the chil 04 Dec 2019, 01:20
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Bunuel
In a family with 3 children, the parents have agreed to bring the children to the pet store and allow each child to choose a pet. This pet store sells only dogs, cats, and monkeys. If each child chooses exactly one animal, and if more than one child can choose the same kind of animal, how many different arrangements of animals could the family leave with?

A. 6
B. 8
C. 9
D. 10
E. 12


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Explanation:

Total Cases can be:
MMM + MMC + MMB + DDD + DDC + DDM + CCC + CCD + CCM + MCB

Total Possibilities of arrangements: 10

IMO-D
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In a family with 3 children, the parents have agreed to bring the chil 16 Dec 2019, 01:51
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Not sure why this question doesn't allow permutation.
CCD is different from CDC or DCC.
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In a family with 3 children, the parents have agreed to bring the chil 16 Dec 2019, 03:36
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sharathnair14
Not sure why this question doesn't allow permutation.
CCD is different from CDC or DCC.
chetan2u
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Hi,

Arrangements could be combination or permutation depending on the context..

If say we were looking for arrangements in which the children could pick the pets, we could go with permutations 3*3*3.
But here we are looking at the arrangements in which the family can leave with. When picked up, all 3 pets are just a group and that group is the arrangement, so combinations is what we are looking for.

But in GMAT, I would expect no ambiguity, and the question should be worded accordingly.
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In a family with 3 children, the parents have agreed to bring the chil 16 Dec 2019, 03:50
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chetan2u
sharathnair14
Not sure why this question doesn't allow permutation.
CCD is different from CDC or DCC.
chetan2u


Hi,

Arrangements could be combination or permutation depending on the context..

If say we were looking for arrangements in which the children could pick the pets, we could go with permutations 3*3*3.
But here we are looking at the arrangements in which the family can leave with. When picked up, all 3 pets are just a group and that group is the arrangement, so combinations is what we are looking for.
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If each child chooses exactly one animal, and if more than one child can choose the same kind of animal, how many different arrangements of animals could the family leave with?
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Hey, chetan2u Thanks and I totally understand your reasoning.
But doesn't this question explicitly state "different arrangements"? That is what caused this doubt.
However, I think now that the difference would be understood when one reads "how many different arrangements of animals" together. Fine margins. Maybe this question could have been a lot more difficult if one of the choices was a permutation product.
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In a family with 3 children, the parents have agreed to bring the chil 02 Mar 2020, 09:59
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Bunuel
In a family with 3 children, the parents have agreed to bring the children to the pet store and allow each child to choose a pet. This pet store sells only dogs, cats, and monkeys. If each child chooses exactly one animal, and if more than one child can choose the same kind of animal, how many different arrangements of animals could the family leave with?

A. 6
B. 8
C. 9
D. 10
E. 12


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We're looking at the ways the family can have "combos" of pets.

3 cases:
1. 3 pets of different kinds. 3! = 6 arrangements
2. 2 pets of a kind. 3!/2! = 3 arrangement.
3. 3 pets of a same kind. 3!/3! = 1 arrangement.

IMO 6+3+1 = 10. C
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In a family with 3 children, the parents have agreed to bring the chil 29 Jan 2024, 09:27
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Bunuel
In a family with 3 children, the parents have agreed to bring the children to the pet store and allow each child to choose a pet. This pet store sells only dogs, cats, and monkeys. If each child chooses exactly one animal, and if more than one child can choose the same kind of animal, how many different arrangements of animals could the family leave with?

A. 6
B. 8
C. 9
D. 10
E. 12


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Solution 1:
Possible arrangements are (DDD) (DDC) (DCC) (DDM) (DMM) (CCC) (MMM) (DCM) (MMC) (CCM) - 10 ways.

Solution 2:
1. One kid chooses D, other chooses C, another chooses M - 3!=6 ways.
2. One kid chooses one kind, another two kids choose same type of pets - 3!/2!=3 ways.
3. All the kids choose same pet - 3!/3!=1 way.
Total: 6+3+1=10 ways.
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In a family with 3 children, the parents have agreed to bring the chil 13 Sep 2025, 04:01
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samarpan.g28


Solution 1:
Possible arrangements are (DDD) (DDC) (DCC) (DDM) (DMM) (CCC) (MMM) (DCM) (MMC) (CCM) - 10 ways.

Solution 2:
1. One kid chooses D, other chooses C, another chooses M - 3!=6 ways.
2. One kid chooses one kind, another two kids choose same type of pets - 3!/2!=3 ways.
3. All the kids choose same pet - 3!/3!=1 way.
Total: 6+3+1=10 ways.
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I think your solution 2 is somewhat contradictory with the cases you laid out in solution 1, for step1 in solution 2 where each kid chooses one there is only one way DCM and not 3! as that would count DCM,MCD etc and other arrangements. For step 2 in solution 2 where two are same and one different over there there's 3 ways to choose the first animal, 1 way to choose the second as two are same and then 2 ways to choose the 3rd and hence 6 ways and this accounts for all the arrangements where two are same in your solution set DDC, DDM etc. Well step three is straight forward as it could be DDD , MMM, CCC.

Bunuel is it the correct approach?
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