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Math Help - Wendy's hamburgers! help urgent

  1. #1
    Newbie HYP3RS0N1C_0V3RDR1VE's Avatar
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    Question Wendy's hamburgers! help urgent

    heres the problem my math teacher gave out. its due 2morrow. 2/29/08

    CHEESE, MAYO, KETCHUP, MUSTARD, PICKLES, ONIONS, TOMATOES, LETTUCE. How many ways can you order a hamburger at Wendy's.

    8th grade math. I know its easy but how can you do the problem without doing the big tree that takes up a lot of space on your paper.
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  2. #2
    Newbie HYP3RS0N1C_0V3RDR1VE's Avatar
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    :(

    come on peoplle i know u are out there.
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  3. #3
    Eater of Worlds
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    Why yes, there is a way.

    Suppose you wanted to order a hamburger with all of those toppings. There is one way to get all 8 toppings. How many ways for two topping?. C(8,2)

    Choosing 3 toppings, C(8,3). And so on.

    How about no toppings?.

    So, we have:

    \sum_{k=0}^{8}C(8,k)

    Can you count them up?. Are you familiar with combinations?.
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  4. #4
    Newbie HYP3RS0N1C_0V3RDR1VE's Avatar
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    we did combinations in 6th grade but i forget how to them because that was like a year and a half ago......
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  5. #5
    Member Henderson's Avatar
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    Another option would be to say:

    Your burger can be either with cheese or no cheese. Two options.
    Your burger can be either with mayo or no mayo. Two options.
    Your burger can be either with ketchup or no ketchup. Two options.
    And so on through all 8 choices, so the total is:

    2\times 2\times 2\times 2\times 2\times 2\times 2\times 2, or 2^8.
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  6. #6
    Newbie HYP3RS0N1C_0V3RDR1VE's Avatar
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    Quote Originally Posted by Henderson View Post
    Another option would be to say:

    Your burger can be either with cheese or no cheese. Two options.
    Your burger can be either with mayo or no mayo. Two options.
    Your burger can be either with ketchup or no ketchup. Two options.
    And so on through all 8 choices, so the total is:

    2\times 2\times 2\times 2\times 2\times 2\times 2\times 2, or 2^8.
    thank you. i will try that.
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  7. #7
    Eater of Worlds
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    Yes, Henderson beat me. I was going to point out the identity:

    2^{n}=\begin{pmatrix}n\\0\end{pmatrix}+\begin{pmat  rix}n\\1\end{pmatrix}+\begin{pmatrix}n\\2\end{pmat  rix}+............+\begin{pmatrix}n\\n\end{pmatrix}

    In your case, n=8

    It may come in handy if you have to sum up combinations.

    I don't what to get too technical. What grade are you in?. What class is this for?. May I ask?.
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  8. #8
    Newbie HYP3RS0N1C_0V3RDR1VE's Avatar
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    7th pre-algebra....
    ur not gonna stalk me are you? lol
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  9. #9
    Eater of Worlds
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    haha, no, I just wanted to get an idea rather I should post the formula or not.
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  10. #10
    Newbie HYP3RS0N1C_0V3RDR1VE's Avatar
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    if u think i could handle it. and if it helps.
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  11. #11
    Newbie HYP3RS0N1C_0V3RDR1VE's Avatar
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    answer?

    i tried galactus's solution and ended up with 288 ways. I also tried henderson's way and got 256 ways. which is correct? if i did do them correct.
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  12. #12
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    It should be 256. 2^8=256.

    With each topping, you can choose to include it or not include it. Thus, you have 2 choices with each topping, so you multiply 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2.
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  13. #13
    Newbie HYP3RS0N1C_0V3RDR1VE's Avatar
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    Quote Originally Posted by Mathnasium View Post
    It should be 256. 2^8=256.

    With each topping, you can choose to include it or not include it. Thus, you have 2 choices with each topping, so you multiply 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2.
    thank you. i was debating over that for an hour now. hehe
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  14. #14
    Eater of Worlds
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    You should get 256 with either way. They're the same. But 2^8 is certainly easier than counting up the combinations from 0 to 8. I was just showing the identity.
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