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Math Help - Probability and counting problems

  1. #1
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    Smile Probability and counting problems

    I need help to solve these problems :

    1. A dartboard is divided into 6 sections. These 6 sections is given numbers : 49, 5, 54, 28, 37, 23. What is the least number of throws needed in order to get a score of exactly 100?
    I have noticed that a sum of 100 can be from the sum of 3 numbers : 23+23+54, 49+28+23. But I have no idea how to work it out.

    2. In a party, there are 2006 cakes. The first group eats 1/2 of them. The second group eats 1/3 of the rest. The third group eats 1/4 of the rest. An so on .... The 2005th group eats 1/2006 of the rest. How many cakes have been eaten ?

    Please help.
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  2. #2
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    Hello, cdesiani!

    You've solved the first one . . .


    1. A dartboard is divided into 6 sections.
    These 6 sections is given numbers: 49, 5, 54, 28, 37, 23.
    What is the least number of throws needed in order to get a score of exactly 100?

    I have noticed that a sum of 100 can be the sum of 3 numbers: 23 + 23 + 54.
    . .
    So you found a three-number sum . . . Good!
    A sum of 100 cannot be produced by one throw.
    And it can be shown that a sum of 100 cannot be produced by two throws.
    . .
    [Just check out the 15 two-number sums.]

    Therefore, at least three throws are needed . . . That's it!

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  3. #3
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    2) None, we have;

    r~=~\left(\frac{1}{2!}\right)+\left(\frac{1}{3!}\r  ight)+\left(\frac{1}{4!}\right)+ ...... +\left(\frac{1}{n!}\right)

    As n\rightarrow\infty~~r\rightarrow0.7182....

    I think
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  4. #4
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    Hello, Sean!

    You are on the right track . . .

    . . r \:= \: \frac{1}{2!}+ \frac{1}{3!}+\frac{1}{4!}+ \hdots + \frac{1}{2006!}


    We have the infinite series: . e^x\;=\;1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \hdots

    . . If x = 1, we have: . e \;=\;1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \hdots

    . . Then:. . e - 2 \;=\;\frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \hdots

    Hence: .  r \;\approx\;e-2 \;=\;0.718281828459045\hdots


    Therefore: . 2006(e-2) \;\approx\;1440\frac{7}{8} cakes were eaten.

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  5. #5
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    Smile confused ....

    Sorry, but what is e that is equal to 0.7182818284...?

    Still
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