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Clarification needed on combinatorics problem

This topic has 4 expert replies and 2 member replies
knight247 Legendary Member
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Clarification needed on combinatorics problem

Post Fri Nov 21, 2014 10:03 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    In how many ways can 16 different gifts be divided among four children such that each child receives exactly four gifts?

    OA 16! ÷ (4!)^4


    I already got the answer by doing 16C4 * 12C4 * 8C4 * 4C4

    My question is, since the four children are NOT identical, shouldn't the above calculation also have a 4C1*3C1*2C1 in there? Considering the four children are NOT identical, we would need to pick one kid each time we need to assign a kid an assortment of four gifts, don't we?

    Detailed explanations would be appreciated. Many thanks in advance.

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    Post Fri Nov 21, 2014 10:14 am
    knight247 wrote:
    In how many ways can 16 different gifts be divided among four children such that each child receives exactly four gifts?

    I already got the answer by doing 16C4 * 12C4 * 8C4 * 4C4.
    Your solution is correct.
    Let the four children be Adam, Bobby, Cindy and David.
    From 16 gifts, the number of ways to choose 4 to give to Adam = 16C4.
    From the remaining 12 gifts, the number of ways to choose 4 to give to Bobby = 12C4.
    From the remaining 8 gifts, the number of ways to choose 4 to give to Cindy = 8C4.
    From the remaining 4 gifts, the number of ways to choose 4 to give to David = 4C4.
    To combine these options, we multiply:
    16C4 * 12C4 * 8C4 * 4C4.

    _________________
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    GMAT Private Tutor
    GMATGuruNY@gmail.com
    If you find one of my posts helpful, please take a moment to click on the "Thank" icon.
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    Last edited by GMATGuruNY on Thu Oct 15, 2015 11:11 am; edited 1 time in total

    Thanked by: knight247
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    Post Fri Nov 21, 2014 10:17 am
    knight247 wrote:
    In how many ways can 16 different gifts be divided among four children such that each child receives exactly four gifts?
    If we take the task of distributing the 16 gifts and break it into [b]stages, we can see that we need not perform the additional calculations you suggest.

    Let's say the children are named A, B, C, and D

    Stage 1: Select 4 gifts to give to child A
    Since the order in which we select the 4 gifts does not matter, we can use combinations.
    We can select 4 gifts from 16 gifts in 16C4 ways (= 16!/(4!)(12!))
    So, we can complete stage 1 in 16!/(4!)(12!) ways

    Stage 2: select 4 gifts to give to child B
    There are now 12 gifts remaining
    Since the order in which we select the 4 gifts does not matter, we can use combinations.
    We can select 4 gifts from 12 gifts in 12C4 ways (= 12!/(4!)(8!))
    So, we can complete stage 2 in 12!/(4!)(8!) ways


    Stage 3: select 4 gifts to give to child C
    There are now 8 gifts remaining
    We can select 4 gifts from 8 gifts in 8C4 ways (= 8!/(4!)(4!))
    So, we can complete stage 3 in 8!/(4!)(4!) ways

    Stage 4: select 4 gifts to give to child C
    There are now 4 gifts remaining
    NOTE: There's only 1 way to select 4 gifts from 4 gifts, but if we want the answer to look like the official answer, let's do the following:
    We can select 4 gifts from 4 gifts in 4C4 ways (= 4!/4!)
    So, we can complete stage 4 in 4!/4! ways

    By the Fundamental Counting Principle (FCP), we can complete all 4 stages (and thus distribute all 16 gifts) in [16!/(4!)(12!)][12!/(4!)(8!)][8!/(4!)(4!)][4!/4!] ways

    A BUNCH of terms cancel out to give us (= 16!/(4!)⁴)

    --------------------------

    Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. For more information about the FCP, watch our free video: http://www.gmatprepnow.com/module/gmat-counting?id=775

    Then you can try solving the following questions:

    EASY
    - http://www.beatthegmat.com/what-should-be-the-answer-t267256.html
    - http://www.beatthegmat.com/counting-problem-company-recruitment-t244302.html
    - http://www.beatthegmat.com/picking-a-5-digit-code-with-an-odd-middle-digit-t273110.html
    - http://www.beatthegmat.com/permutation-combination-simple-one-t257412.html
    - http://www.beatthegmat.com/simple-one-t270061.html
    - http://www.beatthegmat.com/mouse-pellets-t274303.html


    MEDIUM
    - http://www.beatthegmat.com/combinatorics-solution-explanation-t273194.html
    - http://www.beatthegmat.com/arabian-horses-good-one-t150703.html
    - http://www.beatthegmat.com/sub-sets-probability-t273337.html
    - http://www.beatthegmat.com/combinatorics-problem-t273180.html
    - http://www.beatthegmat.com/digits-numbers-t270127.html
    - http://www.beatthegmat.com/doubt-on-separator-method-t271047.html
    - http://www.beatthegmat.com/combinatorics-problem-t267079.html


    DIFFICULT
    - http://www.beatthegmat.com/wonderful-p-c-ques-t271001.html
    - http://www.beatthegmat.com/ps-counting-t273659.html
    - http://www.beatthegmat.com/permutation-and-combination-t273915.html
    - http://www.beatthegmat.com/please-solve-this-real-gmat-quant-question-t271499.html
    - http://www.beatthegmat.com/no-two-ladies-sit-together-t275661.html
    - http://www.beatthegmat.com/laniera-s-construction-company-is-offering-home-buyers-a-wi-t215764.html

    Cheers,
    Brent

    _________________
    Brent Hanneson – Founder of GMATPrepNow.com
    Use our video course along with Beat The GMAT's free 60-Day Study Guide

    Check out the online reviews of our course
    Come see all of our free resources

    Thanked by: knight247
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    prada Senior | Next Rank: 100 Posts Default Avatar
    Joined
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    Posted:
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    1 times
    Post Thu Oct 15, 2015 11:01 am
    GMATGuruNY wrote:
    knight247 wrote:
    In how many ways can 16 different gifts be divided among four children such that each child receives exactly four gifts?

    I already got the answer by doing 16C4 * 12C4 * 8C4 * 4C4.
    Your solution is correct.
    Let the four children be Adam, Bobby, Cindy and David.
    From 16 gifts, the number of ways to choose 4 to give to Adam = 16C4.
    From the remaining 12 gifts, the number of ways to choose 4 to give to Bobby = 12C4.
    From the remaining 8 gifts, the number of ways to choose 4 to give to Cindy = 8C4.
    From the remaining 4 gifts, the number of ways to choose 4 to give to David = 4C4.
    To combine these options, we multiply:
    16C4 * 12C5 * 8C4 * 4C4.
    Hey Mitch on your last line I believe you mean 12C4 and not 12C5?

    Post Thu Oct 15, 2015 11:14 am
    prada wrote:
    Hey Mitch on your last line I believe you mean 12C4 and not 12C5?
    Thanks for pointing out the typo.
    I've amended my solution accordingly.

    _________________
    Mitch Hunt
    GMAT Private Tutor
    GMATGuruNY@gmail.com
    If you find one of my posts helpful, please take a moment to click on the "Thank" icon.
    Available for tutoring in NYC and long-distance.
    For more information, please email me at GMATGuruNY@gmail.com.

    Free GMAT Practice Test How can you improve your test score if you don't know your baseline score? Take a free online practice exam. Get started on achieving your dream score today! Sign up now.
    Dutta Newbie | Next Rank: 10 Posts Default Avatar
    Joined
    31 Jan 2016
    Posted:
    2 messages
    Post Mon Feb 08, 2016 9:54 am
    Hey Brent,

    Since the 4 kids are not identical should we not consider selecting as to who receives the 1st set of 4 gifts and who the 2nd and so on. So shouldn't the ans be multiplied by a 4!?


    Brent@GMATPrepNow wrote:
    knight247 wrote:
    In how many ways can 16 different gifts be divided among four children such that each child receives exactly four gifts?
    If we take the task of distributing the 16 gifts and break it into [b]stages, we can see that we need not perform the additional calculations you suggest.

    Let's say the children are named A, B, C, and D

    Stage 1: Select 4 gifts to give to child A
    Since the order in which we select the 4 gifts does not matter, we can use combinations.
    We can select 4 gifts from 16 gifts in 16C4 ways (= 16!/(4!)(12!))
    So, we can complete stage 1 in 16!/(4!)(12!) ways

    Stage 2: select 4 gifts to give to child B
    There are now 12 gifts remaining
    Since the order in which we select the 4 gifts does not matter, we can use combinations.
    We can select 4 gifts from 12 gifts in 12C4 ways (= 12!/(4!)(8!))
    So, we can complete stage 2 in 12!/(4!)(8!) ways


    Stage 3: select 4 gifts to give to child C
    There are now 8 gifts remaining
    We can select 4 gifts from 8 gifts in 8C4 ways (= 8!/(4!)(4!))
    So, we can complete stage 3 in 8!/(4!)(4!) ways

    Stage 4: select 4 gifts to give to child C
    There are now 4 gifts remaining
    NOTE: There's only 1 way to select 4 gifts from 4 gifts, but if we want the answer to look like the official answer, let's do the following:
    We can select 4 gifts from 4 gifts in 4C4 ways (= 4!/4!)
    So, we can complete stage 4 in 4!/4! ways

    By the Fundamental Counting Principle (FCP), we can complete all 4 stages (and thus distribute all 16 gifts) in [16!/(4!)(12!)][12!/(4!)(8!)][8!/(4!)(4!)][4!/4!] ways

    A BUNCH of terms cancel out to give us (= 16!/(4!)⁴)

    --------------------------

    Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. For more information about the FCP, watch our free video: http://www.gmatprepnow.com/module/gmat-counting?id=775

    Then you can try solving the following questions:

    EASY
    - http://www.beatthegmat.com/what-should-be-the-answer-t267256.html
    - http://www.beatthegmat.com/counting-problem-company-recruitment-t244302.html
    - http://www.beatthegmat.com/picking-a-5-digit-code-with-an-odd-middle-digit-t273110.html
    - http://www.beatthegmat.com/permutation-combination-simple-one-t257412.html
    - http://www.beatthegmat.com/simple-one-t270061.html
    - http://www.beatthegmat.com/mouse-pellets-t274303.html


    MEDIUM
    - http://www.beatthegmat.com/combinatorics-solution-explanation-t273194.html
    - http://www.beatthegmat.com/arabian-horses-good-one-t150703.html
    - http://www.beatthegmat.com/sub-sets-probability-t273337.html
    - http://www.beatthegmat.com/combinatorics-problem-t273180.html
    - http://www.beatthegmat.com/digits-numbers-t270127.html
    - http://www.beatthegmat.com/doubt-on-separator-method-t271047.html
    - http://www.beatthegmat.com/combinatorics-problem-t267079.html


    DIFFICULT
    - http://www.beatthegmat.com/wonderful-p-c-ques-t271001.html
    - http://www.beatthegmat.com/ps-counting-t273659.html
    - http://www.beatthegmat.com/permutation-and-combination-t273915.html
    - http://www.beatthegmat.com/please-solve-this-real-gmat-quant-question-t271499.html
    - http://www.beatthegmat.com/no-two-ladies-sit-together-t275661.html
    - http://www.beatthegmat.com/laniera-s-construction-company-is-offering-home-buyers-a-wi-t215764.html

    Cheers,
    Brent

    Post Mon Feb 08, 2016 1:50 pm
    Dutta wrote:
    Hey Brent,

    Since the 4 kids are not identical should we not consider selecting as to who receives the 1st set of 4 gifts and who the 2nd and so on. So shouldn't the ans be multiplied by a 4!?
    We have already accounted for the children being non-identical.
    At each stage, we give gifts to a particular child (child A, B, C and D)

    Cheers,
    Brent

    _________________
    Brent Hanneson – Founder of GMATPrepNow.com
    Use our video course along with Beat The GMAT's free 60-Day Study Guide

    Check out the online reviews of our course
    Come see all of our free resources

    GMAT Prep Now's comprehensive video course can be used in conjunction with Beat The GMAT’s FREE 60-Day Study Guide and reach your target score in 2 months!

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