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Chemicals color coding system

This topic has 3 member replies
LeoBen Senior | Next Rank: 100 Posts Default Avatar
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Chemicals color coding system

Post Wed Jan 04, 2012 4:28 pm
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    In a certain lab, chemicals are identified by a color coding system. There are 20 different chemicals. Each is coded with either a single color or a unique two color combination. If the order of colors in the pairs doesnt matter, what is the miniumum number of different colors needed to code all 20 chemicals with either a single color or a unique pair of color.

    a. 5
    b. 6
    c. 7
    d. 20
    e. 40

    OA: B

    My answer doesnt equate OA, below is my approach (backsolving) PLEASE correct me. Thanks.

    consider a. 5 as it is the smallest amongst all.

    1 or 2 nos = _ + _ _

    5 + 5C2 X 2! = 5 + 10 X 2! = 25 --> this is sufficient as other will give me more than this value.

    Checking with the OE - the only contention is the 2! in 2 colors, my rationale being the 2 colors chosen can be arranged in two ways. For ex. we choose red & blue - these can be arranged as: RB or BR.
    Analogy I thought - if we have 3 colors RAG (red, amber, green) , selecting 2 of them we can form 6 of them:
    RA
    AG
    GR
    AR
    GA
    RG
    ------------------------- Am I missing a point here?

    Source: Kaplan

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    pemdas Legendary Member Default Avatar
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    Post Wed Jan 04, 2012 5:45 pm
    this one is quite easy without backsolving

    I'll try to explain the logic firstly

    you can select two color sets without order priority among n colors -> nC2
    your selection must return at most 20 - also your selection may return less than 20 (you need to trade-off here Smile in the solution process) -> nC2 =< 20
    find such value for n so that n distributed in two-color sets Plus some value of (n-value) is enough to cover up 20 chemicals

    Let's see, nC2 and n=7, hmm we have nC2=21 (greater than 20) reject
    nC2 and n=6, we get nC2=15, OK fine, we also have 5 colors among six individual colors which are distributed in pairs. Thus, total makes 15 two-color sets made up out of six colors + 5 solid colors out of six colors makes minimum 6 colors (revised for clarity)

    b

    your mistake was in permuting nP2 instead of nC2, when the problem explicitly specifies the order in pairs doesn't matter. In your way, 5P2 is 20 and you get there ans.A which is wrong BTW.
    LeoBen wrote:
    In a certain lab, chemicals are identified by a color coding system. There are 20 different chemicals. Each is coded with either a single color or a unique two color combination. If the order of colors in the pairs doesnt matter, what is the miniumum number of different colors needed to code all 20 chemicals with either a single color or a unique pair of color.

    a. 5
    b. 6
    c. 7
    d. 20
    e. 40

    OA: B

    My answer doesnt equate OA, below is my approach (backsolving) PLEASE correct me. Thanks.

    consider a. 5 as it is the smallest amongst all.

    1 or 2 nos = _ + _ _

    5 + 5C2 X 2! = 5 + 10 X 2! = 25 --> this is sufficient as other will give me more than this value.

    Checking with the OE - the only contention is the 2! in 2 colors, my rationale being the 2 colors chosen can be arranged in two ways. For ex. we choose red & blue - these can be arranged as: RB or BR.
    Analogy I thought - if we have 3 colors RAG (red, amber, green) , selecting 2 of them we can form 6 of them:
    RA
    AG
    GR
    AR
    GA
    RG
    ------------------------- Am I missing a point here?

    Source: Kaplan

    _________________
    Success doesn't come overnight!

    Thanked by: LeoBen
    LeoBen Senior | Next Rank: 100 Posts Default Avatar
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    Post Fri Jan 06, 2012 12:02 am
    Oh yes - my mistake. Ofcourse order doesnt matter, so RB or BR its just the same as per the stem i missed it I guess.

    Thanks Pemdas

    ArunangsuSahu Master | Next Rank: 500 Posts Default Avatar
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    Post Fri Jan 06, 2012 1:01 am
    Knowledge of Combination

    The problem is same as Either 1 or 2 choices...


    6C1+6C2=21>20

    So minimum 6 Colors needed

    (B) is the answer

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