GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 19 Nov 2018, 01:41

INSEAD R1 Results This Week!

First Decision Reported on Decision Tracker  |  Join INSEAD Chat to Calm Your Nerves & Catch the Latest Action


Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel
Events & Promotions in November
PrevNext
SuMoTuWeThFrSa
28293031123
45678910
11121314151617
18192021222324
2526272829301
Open Detailed Calendar
  • How to QUICKLY Solve GMAT Questions - GMAT Club Chat

     November 20, 2018

     November 20, 2018

     09:00 AM PST

     10:00 AM PST

    The reward for signing up with the registration form and attending the chat is: 6 free examPAL quizzes to practice your new skills after the chat.
  • The winning strategy for 700+ on the GMAT

     November 20, 2018

     November 20, 2018

     06:00 PM EST

     07:00 PM EST

    What people who reach the high 700's do differently? We're going to share insights, tips and strategies from data we collected on over 50,000 students who used examPAL.

In a family with 3 children, the parents have agreed to brin

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Intern
Intern
avatar
Joined: 05 May 2010
Posts: 37
In a family with 3 children, the parents have agreed to brin  [#permalink]

Show Tags

New post 18 Aug 2010, 13:41
2
13
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

43% (01:46) correct 57% (01:26) wrong based on 253 sessions

HideShow timer Statistics

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
Most Helpful Community Reply
Director
Director
avatar
Status: Apply - Last Chance
Affiliations: IIT, Purdue, PhD, TauBetaPi
Joined: 17 Jul 2010
Posts: 628
Schools: Wharton, Sloan, Chicago, Haas
WE 1: 8 years in Oil&Gas
Re: Combinatorics [McGraw Hill's GMAT - 2010]  [#permalink]

Show Tags

New post 18 Aug 2010, 15:02
6
1
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
_________________

Consider kudos, they are good for health

General Discussion
Intern
Intern
avatar
Joined: 05 May 2010
Posts: 37
Re: Combinatorics [McGraw Hill's GMAT - 2010]  [#permalink]

Show Tags

New post 18 Aug 2010, 15:31
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?
Director
Director
avatar
Status: Apply - Last Chance
Affiliations: IIT, Purdue, PhD, TauBetaPi
Joined: 17 Jul 2010
Posts: 628
Schools: Wharton, Sloan, Chicago, Haas
WE 1: 8 years in Oil&Gas
Re: Combinatorics [McGraw Hill's GMAT - 2010]  [#permalink]

Show Tags

New post 18 Aug 2010, 16:09
2
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
_________________

Consider kudos, they are good for health

Intern
Intern
avatar
Joined: 05 May 2010
Posts: 37
Re: Combinatorics [McGraw Hill's GMAT - 2010]  [#permalink]

Show Tags

New post 19 Aug 2010, 11:41
Ah excellent Mainhoon, thank you very much :); that is exactly the approach I was trying to figure out. Your nCk explanation is perfect!
Retired Moderator
User avatar
Status: 2000 posts! I don't know whether I should feel great or sad about it! LOL
Joined: 04 Oct 2009
Posts: 1194
Location: Peru
Schools: Harvard, Stanford, Wharton, MIT & HKS (Government)
WE 1: Economic research
WE 2: Banking
WE 3: Government: Foreign Trade and SMEs
Re: Combinatorics [McGraw Hill's GMAT - 2010]  [#permalink]

Show Tags

New post 25 Aug 2010, 11:16
mainhoon wrote:
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!
_________________

"Life’s battle doesn’t always go to stronger or faster men; but sooner or later the man who wins is the one who thinks he can."

My Integrated Reasoning Logbook / Diary: http://gmatclub.com/forum/my-ir-logbook-diary-133264.html

GMAT Club Premium Membership - big benefits and savings

Intern
Intern
avatar
Joined: 10 Oct 2010
Posts: 21
Location: Texas
Re: Combinatorics [McGraw Hill's GMAT - 2010]  [#permalink]

Show Tags

New post 15 Oct 2010, 00:01
1
BigBrad wrote:
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!

Solution:
5 Options
3 Identical "Picks"
2 Identical "Skips"
5! / (3! * 2!) = 10

ANS: D
Intern
Intern
avatar
Joined: 01 Aug 2010
Posts: 6
Re: Combinatorics [McGraw Hill's GMAT - 2010]  [#permalink]

Show Tags

New post 28 Feb 2011, 06:57
1
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,.....
Intern
Intern
avatar
Joined: 21 Oct 2011
Posts: 4
Re: Combinatorics [McGraw Hill's GMAT - 2010]  [#permalink]

Show Tags

New post 09 Nov 2011, 04:54
excellent analysis puneet.... I got it now.... thanks for explaining in detail.
Intern
Intern
avatar
Joined: 21 Oct 2011
Posts: 4
Re: Combinatorics [McGraw Hill's GMAT - 2010]  [#permalink]

Show Tags

New post 09 Nov 2011, 04:54
excellent analysis puneet.... I got it now.... thanks for explaining in detail.
Intern
Intern
avatar
Joined: 03 Dec 2012
Posts: 4
GPA: 3.64
Re: Combinatorics [McGraw Hill's GMAT - 2010]  [#permalink]

Show Tags

New post 16 Jan 2013, 09:26
1
Does my solution make sense?

We have these possibilities:

1) The three children choose three different animals: \(3C3 = 1\)

2) The three children all choose the same type of animal: \(3C1 = 3\)

3) Two of the three children pick the same type of animal whereas one child picks another animal: \(3C2*2! = 3*2! = 6\)

Now, \(1+3+6 = 10\)
Manager
Manager
avatar
Joined: 18 Oct 2011
Posts: 87
Location: United States
Concentration: Entrepreneurship, Marketing
GMAT Date: 01-30-2013
GPA: 3.3
Re: In a family with 3 children, the parents have agreed to brin  [#permalink]

Show Tags

New post 16 Jan 2013, 13:20
Visualize arragements...order doesnt matter here so we just need combos.

DDD,CCC,MMM (all 3)
DDC, DDM, CCM, CCD, MMD, MMC (2 and 1)
D,C,M (only 1 animal each)

= 10 (D)
Senior Manager
Senior Manager
User avatar
G
Joined: 03 Apr 2013
Posts: 278
Location: India
Concentration: Marketing, Finance
GMAT 1: 740 Q50 V41
GPA: 3
GMAT ToolKit User
In a family with 3 children, the parents have agreed to brin  [#permalink]

Show Tags

New post 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.
_________________

Spread some love..Like = +1 Kudos :)

Current Student
User avatar
B
Status: DONE!
Joined: 05 Sep 2016
Posts: 378
Re: In a family with 3 children, the parents have agreed to brin  [#permalink]

Show Tags

New post 27 Nov 2016, 11:23
AAA
BBB
CCC

(all one kind)

AAB
AAC
BBA
BBC
CCA
CCB

(two of a kind)

ABC

(each selected)

10 combinations
Senior Manager
Senior Manager
User avatar
G
Joined: 03 Apr 2013
Posts: 278
Location: India
Concentration: Marketing, Finance
GMAT 1: 740 Q50 V41
GPA: 3
GMAT ToolKit User
Re: In a family with 3 children, the parents have agreed to brin  [#permalink]

Show Tags

New post 19 Jun 2017, 00:54
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 that the question is wrong in that it does not specify in exactness what it is asking for.

As has been said in the question, that there are 3 types of animals, and they have asked us for the number of different permutations that the family can leave with, we will have the following cases.

1. All same
There are 3 cases here
ddd, ccc, and mmm.

But the question treats all these as 1. This is wrong.

The same goes for the cases

cdm

and

ccm, mmc, ddc, ccd,....

Hi Bunuel, please take a look.
_________________

Spread some love..Like = +1 Kudos :)

Manager
Manager
avatar
S
Joined: 27 Aug 2014
Posts: 56
Location: Canada
Concentration: Strategy, Technology
GMAT 1: 660 Q45 V35
GPA: 3.66
WE: Consulting (Consulting)
Premium Member
Re: In a family with 3 children, the parents have agreed to brin  [#permalink]

Show Tags

New post 21 Jun 2017, 22:30
Since the options were all small numbers I just basically counted

MMM
MMC
MMD

CCC
CCD
CCM

DDD
DDC
DDM

MDC

so 10 ways
Non-Human User
User avatar
Joined: 09 Sep 2013
Posts: 8821
Premium Member
Re: In a family with 3 children, the parents have agreed to brin  [#permalink]

Show Tags

New post 18 Oct 2018, 03:03
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).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

GMAT Books | GMAT Club Tests | Best Prices on GMAT Courses | GMAT Mobile App | Math Resources | Verbal Resources

GMAT Club Bot
Re: In a family with 3 children, the parents have agreed to brin &nbs [#permalink] 18 Oct 2018, 03:03
Display posts from previous: Sort by

In a family with 3 children, the parents have agreed to brin

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  


Copyright

GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.