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    myhrerd's Avatar
    myhrerd Posts: 1, Reputation: 1
    New Member
     
    #1

    Jan 9, 2008, 08:24 PM
    probability of combinations
    If you are building hamburgers and have 8 different toppings, how many different combinations of hamburgers can you make?
    twinkiedooter's Avatar
    twinkiedooter Posts: 12,172, Reputation: 1054
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    #2

    Jan 9, 2008, 08:46 PM
    I haven't been to Burger King in years so I'm not even going to be close with my answer. I only need one hamburger with all 8 toppings on it.
    N0help4u's Avatar
    N0help4u Posts: 19,823, Reputation: 2035
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    #3

    Jan 9, 2008, 08:55 PM
    hamburger, bun and 1 topping each =8
    hamburger, bun and 8 toppings =8
    hamburger, bun and 7 toppings x _ different ways
    hamburger, bun and 6 topping x _ different ways
    hamburger, bun and 5 toppings x _ different ways
    hamburger, bun and 4 toppings x _ different ways
    hamburger, bun and 3 toppings x_different ways
    hamburger, bun ketchup, mustard,
    hamburger, bun ketchup, onion
    hamburger, bun ketchup, relish
    hamburger, bun ketchup, saurkraut
    hamburger, bun ketchup, cheese
    hamburger, bun ketchup, chili
    hamburger, bun ketchup, secret ingredient

    hamburger, bun mustard, onion
    hamburger, bun mustard, relish
    hamburger, bun mustard, saurkraut
    hamburger, bun mustard, cheese
    hamburger, bun mustard, chili
    hamburger, bun mustard, secret ingredient

    hamburger, bun &...

    I got lost :(...

    Plus 0NE plain!
    galactus's Avatar
    galactus Posts: 2,271, Reputation: 282
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    #4

    Jan 10, 2008, 01:33 PM
    The problem is rather ambiguous. I assume you can use any amount of topping. You don't have to use all 8 and order doesn't matter.

    There is C(8,1)=8 ways to choose one topping

    C(8,2) = 28 ways to choose 2 toppings

    C(8,3)=56 ways to choose 3 toppings

    C(8,4)=70 ways to choose 4 toppings

    C(8,5)=56 ways to choose 5 toppngs

    C(8,6)=28 ways to choose 6 toppings

    C(8,7)=8 ways to choose 7 toppings

    C(8,8)=1 way to choose 8 toppings

    Of course, 1 way for no toppings

    Add them all up and we get 256 possible hambuger combos.
    andreafc's Avatar
    andreafc Posts: 10, Reputation: 2
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    #5

    Jan 11, 2008, 02:57 AM
    Galactus is right. There is a general way to get to this result with a simple formula, that works with any number in just one step. However, since I 'm too lazy to post embedded math, I attach here a pdf explaining how.
    Attached Images
  1. File Type: pdf combinations.pdf (24.1 KB, 265 views)

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