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Math Help - Tricky Combination Questions

  1. #1
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    Tricky Combination Questions

    1. Solve the equation for n algebraically.

    nCn-2 = 10

    Would I have to divide the left side by (n)(n)(n-1)(n-2), which would cancel out (n) and (n-2). Im stuck with what to do after that.

    2. A pencil case can hold seven pens and eight pencils. Meg chooses either a pen or pencil. Then John is to choose a pen and a pencil. In which case (based on Andrea's initial choice) would John have to have the greater number of choices.

    I think Meg's choice would be: 7C1 x 8C0 + 7C0 x 8C1
    Then i get stuck on John. Would it be 6C1 x 7C1

    Thanks.


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  2. #2
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    Quote Originally Posted by CroatBoy View Post
    2. A pencil case can hold seven pens and eight pencils. Meg chooses either a pen or pencil. Then John is to choose a pen and a pencil. In which case (based on Andrea's initial choice) would John have to have the greater number of choices.[/SIZE][/SIZE][/SIZE][/SIZE]
    If Meg chooses a pen then the total number of ways is (7)(6)(8).
    If Meg chooses a pencil then the total number of ways is (8)(7)(7).
    No she does either and not both,so add.
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  3. #3
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    Quote Originally Posted by Plato View Post
    If Meg chooses a pen then the total number of ways is (7)(6)(8).
    If Meg chooses a pencil then the total number of ways is (8)(7)(7).
    No she does either and not both,so add.
    I dont understand what mean. Why would you multiply 7x6x8 and 8x7x7?
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  4. #4
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    Quote Originally Posted by CroatBoy View Post
    I dont understand what mean. Why would you multiply 7x6x8 and 8x7x7?
    There are 7 ways for Meg to select a pen. Once that is done the John has 6 ways to select a pen and 8 ways to pick a pencil: (7)(6)(8).
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  5. #5
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    Quote Originally Posted by Plato View Post
    There are 7 ways for Meg to select a pen. Once that is done the John has 6 ways to select a pen and 8 ways to pick a pencil: (7)(6)(8).
    Alright. I see what you mean, thanks alot.
    What about # 1?
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  6. #6
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    Quote Originally Posted by CroatBoy View Post
    Alright. I see what you mean, thanks alot.
    What about # 1?
    So in #1, I'm not sure of the notation, but here's what I think it's saying.

    \frac{n!}{(n-2)! 2!} = 10
    \frac{n(n-1)}{2}=10
    n^2 - n - 20 = 0
    n = -4,5

    Only the positive one makes sense, so pick n=5.
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