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Thread: color coding

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    color coding

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    a company has 15 disrtibution centres and uses color coding to identify each center.either a single color or a pair of two different colors is chosen to represent each center uniquely.what is the mnimum number of colors needed for the coding and the order of the colors doesnot matter?

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    Answer is 5. If it is correct I can provide an explanation.

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    Isnt't this simply a factorial problem?

    Answer should be 4. (I think)

    4! = 4*3*2*1 = 24, and we are using 2 colors, so we divide by 2 = 12.
    12 + the original 4 colors = 16 which will cover the 15 centers.

    A B C D AB AC AD BA BC BD CA CB CD DA DB DC.

    This is how I solved it.

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    rexepic: in the question it is mentioned thet "the order of the colors does not matter" so in the answer provided by you, CA and AC will not be counted distinct.

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    i think masuhara is right.
    if you have 5 different colors then there are
    5 C 2 = 5!/ (3! 2!)= 10
    ways to pick two different colors.So that takes care of 10 distribution centers. Then you can use the 5 colors for the rest of the 5 distribution centers

    So there'll be 10 distribution centers with two colors and 5 with 1 color each

    you cant do it with 4 or less colors

    this is a very hard problem, by the way lol ..

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    The answer is 4 not five.

    Actually, this is a typical problem testing the Binomial Theorem's Pascal's triangle if you remember:

    n = 0, 1
    n = 1, 1 1
    n = 2, 1 2 1
    n = 3, 1 3 3 1
    n = 4, 1 4 6 4 1
    n = 5, 1 5 10 10 5 1

    Here we need to ignore the 1's at the left end and I will explain you why:

    The question says - There are n colors that can be arranged in whatever form as long as arrangements do not matter such that sum of these combinations of colors is >= 15

    Let us assume we need 4 colors therefore we have to find:
    how we can arrange the 4 colors taken 1 at a time, taken 2 at a time, taken 3 at a time and so on:
    i.e.
    4C1 + 4C2 + 4C3 + 4C4 = 4 + 6 + 4 + 1 = 15 -> reason we ignored the first term in the pascal's triangle is because it denotes nC0.

    So the answer should be 4 such colors are required.

    let us look at the options as well - colors = a,b,c,d

    4c1 = a,b,c,d
    4c2 = ab, bc, bd, ac, ad, cd
    4c3 = abc, abd, bcd, acd
    4c4 = abcd
    4+6+4+1 = 15
    Hey Harvard, I am right here!!
    rep me if I made some sense

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    Quote Originally Posted by ram22 View Post
    a company has 15 disrtibution centres and uses color coding to identify each center.either a single color or a pair of two different colors is chosen to represent each center uniquely.what is the mnimum number of colors needed for the coding and the order of the colors doesnot matter?
    Let x represent the number of distinct colors

    # centers covered by one color = x
    # centers covered by two colors = x*(x-1)/2

    So, we want...

    x+x^2/2-x/2 >= 15

    (x^2)/2 + x/2 - 15 >= 0

    x^2 + x - 30 >= 0

    (x + 6) (x - 5) >= 0

    x = 5 colors

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    (I don't why) I sometimes feel insecured with P&C. In that case, I get back to the basics.

    The following way can be helpful (to some of us)
    Write down distribution centers numbers ( as 1. 2. ..3 15) then gives them color (as A.. B...C ...)
    1. A
    2. B
    3.AB
    4. BC (AS ORDER IS NOT MPORTANT TAKE ANOTHER COLOR)
    5. C
    6. CA
    7. CD
    8. D
    9. DA
    10. DB
    11. DE
    12. E
    13. EA
    14. EB
    15. EC

    It is basic, but managable flawlessly
    Cheers !!!
    Dont think about the difficulty, Just get it right !
    If my post is helpful, click "+" button. Lets inspire each other !!!

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    Correct answer is 5, for reasons given (and personally, I'm in favor of using brute force for problems where that will work, including this one.)

    But I did notice something kinda cool: the simplest formula for how many distribution centers can be labeled with a given number of colors (given the terms of the problem) is (n+1)C2. Conceptually, this is how many combinations of two colors can be found, counting "no color" as a color.Fun stuff.

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