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In a family with 3 children, the parents have agreed to brin [#permalink]

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18 Aug 2010, 13:41

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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?

The question has a catch.. It would seem on the face of it it would be 3 x 3 x3 = 27 arrangements, but what it is asking is how many arrangements the family leaves with.. so that is going to be Pick all 3 = 1 way Pick all of a kind = 3 x 1 = 3 ways Pick 2 of a kind = 3 x 1 x 2 = 6 ways Total of 10
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I think the best way to approach the problem (or any problem) is to understand the fundamental - if you adopt a nPk or nCk approach it is bound to fail. Notice that I did use the nCk approach. Here is how: Pick all 3 = Pick one of D C or M = 3C1 = 3 Pick one of each = 1C1 x 1C1 x 1C1 = 1 Pick two of a kind = Pick the kind to repeat x combinations of that kind = 3C1 x 2C1 (two left after you picked the first) = 6

So 10 again.. I don't think there is one formula here that will get you 10, you will need to add the individual combinations.. Perhaps someone else can comment
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This is quite an easy question if done manually but I am having difficulty trying to work it out using the combinatorics formula approach ie nCk. It should be so simple but all my calculations keep producing numbers greater than any of the answer choices. Please could someone show me how to solve this using the formula, thanks.

"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

Question Type: Combination w/ Repetition Technique: Select and Skip

Picture a basket of dogs, a basket of cats, and a basket of monkeys. Let: X mean pick up an animal, and -> mean skip to the next basket.

In order to get 3 animals, the family will have to perform 5 actions. For example: If the family wanted 3 monkeys, their actions would be: -> -> X X X If they want a dog and 2 monkeys: X -> -> X X If they want one of each: X -> X -> X

Now the problem has been simplified to: In how many ways can you arrange the five actions Pick, Pick, Pick, Skip, Skip?

New Question Type: Arrangement (Permutation), No Replacement, Using All Options, Identical Options Technique: Options! / Identical!

Thanks for the explanation Mainhoon, it seems then that this type of question can't actually be solved by any method other than manual brute force i.e. we can't use the nCk combination formula in some way?

The question has a catch.. It would seem on the face of it it would be 3 x 3 x3 = 27 arrangements, but what it is asking is how many arrangements the family leaves with..

Hi Mainhoon, how could you identify that the question is not asking you the typical combination (in this case: 3x3x3)?, could you provide another example to understand it much better?, how differentiate between these 2 types of problem?

Thanks a lot!
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The question is asking about different arrangements that the family may walk out with- we have to choose from 3 different categories of animals and arrange them among 3 children---- ARRANGE is the key word here.

3 ways possible- 1. Each one picks different animal (no. of ways of arranging 3 things all different) 3!=6ways. 2. Each one picks the same-(no. of ways of arranging 3 things all identical) 3!/3!=1 3. 2 same one different-(no. of ways of arranging 3 things, 2 identical) 3!/2!=3 So a total of 6+3+1=10 ways that the family may walk out with......

why not 27? Let the children be numbered 1 2 and 3. No. of ways to select and arrange from among three category of animals Ds Cs Ms, Each dog, cat and monkey is indistinguishable here. If each of the animal from among the category were different for ex- d1, d2 d3 then there would have been 27 ways that the family could have walked out with.........

This is what i could make of the question............correct me if wrong,.....

Re: In a family with 3 children, the parents have agreed to brin [#permalink]

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23 Jul 2014, 02:12

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).

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Re: In a family with 3 children, the parents have agreed to brin [#permalink]

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14 Aug 2016, 18:16

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).

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In a family with 3 children, the parents have agreed to brin [#permalink]

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26 Nov 2016, 21:48

BigBrad wrote:

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

I think the question is flawed. Although the answer for "number of combinations/selections" will be 10, but as the question specifies "arrangements", the answer should be different. 10 should be multiplied with 3! as all the three animals will be distinct. Bunuel Please see.
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