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Math Help - Sandwhich choices

  1. #1
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    Sandwhich choices

    Bobby is about ready to order a sub sandwich from the sandwich shop. The shop offers 4 types of bread, two sizes of sandwhiches, and ten items that he can choose to have on the sandwich. One can get a single or double order of any of these ten items. How many choices does bobby have for consideration? How many choices does he have if he wants to have at least chicken and lettuce on his sandwich?
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    Quote Originally Posted by ihavvaquestion View Post
    Bobby is about ready to order a sub sandwich from the sandwich shop. The shop offers 4 types of bread, two sizes of sandwhiches, and ten items that he can choose to have on the sandwich. One can get a single or double order of any of these ten items. How many choices does bobby have for consideration? How many choices does he have if he wants to have at least chicken and lettuce on his sandwich?
    Have you tried this problem yourself? The first question tests basic understanding of the counting principle. There's little I can do besides explain the counting principle, or simply give you the answer.

    Here's how the principle works. Say you're building a house, and you have two types of roof to choose from, and three types of windows, and that's all you're concerned about. So the number of ways to choose roof and windows is 2*3 = 6.

    You can list them to see why. Say roofs are either A = angled, F = flat. Windows are either O = opaque, T = translucent, C = clear. (Yes, opaque windows are silly, but this is just an example.) Then the 6 possibilities are:

    A, O
    A, T
    A, C
    F, O
    F, T
    F, C

    Please try to apply this principle to the problem. The follow-up problem with chicken and lettuce is a little more involved, but if you don't understand the counting principle it's useless to try to explain anything that builds on it.
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    yeah man, I tried the problem...i was just posting it to see if what I was doing was correct.

    I got:

    4 bread choices * 2 size choices * 2^10 item choices *2 choices for single or double order

    4*2*2*2^10= 16384 total choices

    Then I figured 4*2*2*2^8= 4096 would be all choices without chicken or lettuce so i subtracted that from the total choices to get all choices with chix and lett

    so i got 16384-4096= 12288 choices with chicken and lettuce


    Sometimes if I see someone else work a problem, it helps me learn. Sometimes posting a problem and seeing a solution confirms what I have done is right or wrong and that i was doing it right or wrong.
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    Quote Originally Posted by ihavvaquestion View Post
    yeah man, I tried the problem...i was just posting it to see if what I was doing was correct.

    I got:

    4 bread choices * 2 size choices * 2^10 item choices *2 choices for single or double order

    4*2*2*2^10= 16384 total choices

    Then I figured 4*2*2*2^8= 4096 would be all choices without chicken or lettuce so i subtracted that from the total choices to get all choices with chix and lett

    so i got 16384-4096= 12288 choices with chicken and lettuce


    Sometimes if I see someone else work a problem, it helps me learn. Sometimes posting a problem and seeing a solution confirms what I have done is right or wrong and that i was doing it right or wrong.
    Ah, I wasn't trying to insinuate anything; the main issue is that it's hard to know what area you need help with when all we have is the problem statement. Often it's the case that the poster has no idea how to solve the problem and/or wants other people to do the work for them, we just have no way of knowing.

    For the first problem, the single and double orders seems to have tripped you up. We have ten items, but each of these items has two possible variations total. That means, for any particular item, we can either (1) not have it, (2) have single of it, or (3) have double of it. So it becomes

    4\cdot2\cdot3^{10}

    The way you solved it, Bobby can only get all single orders or all double orders but can't mix and match.

    Now for the second problem, there are four ways to choose how to get the chicken and lettuce: single/single, single/double, double/single, double/double.

    So it becomes 4\cdot2\cdot4\cdot3^{8}

    Make sense?

    (Edited out an error, in case anyone noticed.)
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    ok, i understand the first part is 4*2*3^10

    but for the second part, dont i need to take the answer from number one and subtract out all choices with chicken and lettuce?

    so would it be 4*2*3^10 - 4*2*3^8 ; total combinations minus the combinations without chicken

    i dont understand why it would just become 4*4*2*10^8
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    Quote Originally Posted by ihavvaquestion View Post
    ok, i understand the first part is 4*2*3^10

    but for the second part, dont i need to take the answer from number one and subtract out all choices with chicken and lettuce?

    so would it be 4*2*3^10 - 4*2*3^8 ; total combinations minus the combinations without chicken

    i dont understand why it would just become 4*4*2*10^8
    There's more than one way to solve many problems.

    For my approach, I took: the number of ways to choose chicken/lettuce, multiplied by the number of ways to choose the other elements.

    The solution you wrote doesn't work because 4*2*3^8 is the number of ways to choose sandwiches without chicken or lettuce, but does not include the possibility that you have chicken but not lettuce, or lettuce but not chicken.

    In order to get your way to work, you'll need

    4*2*3^10 - 4*2*3^8 - 4*2*2*3^8 - 4*2*2*3^8

    Where the bold 2's represent choosing either single or double of chicken or lettuce. The answer is the same, but I think my way is easier.
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