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Math Help - permutation

  1. #16
    A Plied Mathematician
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    Still incorrect.
    1 bit: 2 ways.
    2 bits: 4 ways.
    3 bits: 8 ways.
    4 bits: 16 ways.

    What is this pattern?
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  2. #17
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    heheh
    it look like a Geometric Progression
    let me try Sum for S10..hope i will get the ans
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  3. #18
    MHF Contributor Unknown008's Avatar
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    Yes... but the codes of the shop consist of 10 bits only, no less.
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  4. #19
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    Quote Originally Posted by Ackbeet View Post
    Still incorrect.
    1 bit: 2 ways.
    2 bits: 4 ways.
    3 bits: 8 ways.
    4 bits: 16 ways.


    What is this pattern?
    let a =2 ways r=4/2=2

    So, S10= 2[2^10 - 1] per 2-1 = 2046

    huwa still not same
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  5. #20
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    Closer, but still no cigar. What am I doing to each number in this sequence: 2, 4, 8, 16, 32, ... to get the next number? Furthermore, which number in that sequence do you need?
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  6. #21
    MHF Contributor Unknown008's Avatar
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    Quote Originally Posted by nikk View Post
    let a =2 ways r=4/2=2

    So, S10= 2[2^10 - 1] per 2-1 = 2046

    huwa still not same
    Ok, I'll say it another way.

    You are finding for:
    1 bit
    2 bits
    3 bits
    etc

    but how many ways are there for 10 bits?
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  7. #22
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    Quote Originally Posted by Ackbeet View Post
    Closer, but still no cigar. What am I doing to each number in this sequence: 2, 4, 8, 16, 32, ... to get the next number? Furthermore, which number in that sequence do you need?
    the sequense is when n=10. i'm i right prof? hehehe
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  8. #23
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    Yes, you're interested in the 10th term of that sequence. What is that number?
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  9. #24
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    Quote Originally Posted by Ackbeet View Post
    Yes, you're interested in the 10th term of that sequence. What is that number?
    ehheh i get it now...it is T10 = 1024..hahahha
    i get it.

    why not we use Sn, it consider as Sn to get the sum of T1 to T10. it more logic??? it my thinking..heheh
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  10. #25
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    Because if I have 2 ways of doing one thing, and 3 ways of doing another, I don't have 5 ways of doing them both, I have 6. Think of it this way: for each way of doing the first thing, I have 3 ways of doing the second. Therefore, there must be 6 ways to do them both.
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  11. #26
    MHF Contributor Unknown008's Avatar
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    Quote Originally Posted by nikk View Post
    ehheh i get it now...it is T10 = 1024..hahahha
    i get it.

    why not we use Sn, it consider as Sn to get the sum of T1 to T10. it more logic??? it my thinking..heheh
    Because the code of the shop consists of 10 digits. Let's look at it another way.

    _ _ _ _ _ _ _ _ _ _

    Ok, I put 10 blanks. In each blank, you can put 2 digits, that is either 0 or 1.

    In the first blank, you have two choices, in the second, again two choices, in the third, again two choices, etc.

    This gives: 2\times2\times2\times2\times2\times2\times2\times2  \times2\times2 = 2^{10} = 1024
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  12. #27
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    Quote Originally Posted by Unknown008 View Post
    Because the code of the shop consists of 10 digits. Let's look at it another way.

    _ _ _ _ _ _ _ _ _ _

    Ok, I put 10 blanks. In each blank, you can put 2 digits, that is either 0 or 1.

    In the first blank, you have two choices, in the second, again two choices, in the third, again two choices, etc.

    This gives: 2\times2\times2\times2\times2\times2\times2\times2  \times2\times2 = 2^{10} = 1024
    it mean this is a simple permutation. using a box method.but i also like to use Ackbeet method also. it new and relate to another chapter i.e progression. thank for both of you man
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