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Math Help - permutation

  1. #1
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    Unhappy permutation

    hai...i need help for the following basic problem. hope some on can guide me...

    Question:

    Digital devises store information in the form of digits using only the digits 0 or 1. Calculate the number of codes which can be stored if they consist of 10 digits.

    Ans:
    1024

    tq for your advice.
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  2. #2
    A Plied Mathematician
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    Is that your answer? If so, how did you arrive at it?
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  3. #3
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    Quote Originally Posted by Ackbeet View Post
    Is that your answer? If so, how did you arrive at it?
    no..that the answers from book. i can't get it. i need the help in order to get the ans.

    tq for your response.
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  4. #4
    A Plied Mathematician
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    Ok. So what thoughts have you had?
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  5. #5
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    Quote Originally Posted by Ackbeet View Post
    Ok. So what thoughts have you had?
    what should i list..

    is it 0 1 0 1 0 1 0 1 0 1 or 0 1 2 3 4 5 6 7 8 9....??

    that is my big problem...

    huwahuwa
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  6. #6
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    You have ten digits, each of which could be 0 or 1. So one possibility would be 0000000000. Another would be 0000000001. Then you could have 0000000010. And so on. How many ways could you have these bits flipped?
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  7. #7
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    Quote Originally Posted by Ackbeet View Post
    You have ten digits, each of which could be 0 or 1. So one possibility would be 0000000000. Another would be 0000000001. Then you could have 0000000010. And so on. How many ways could you have these bits flipped?
    ok2..let me try..hope u still there

    heheeh
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  8. #8
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    Quote Originally Posted by Ackbeet View Post
    You have ten digits, each of which could be 0 or 1. So one possibility would be 0000000000. Another would be 0000000001. Then you could have 0000000010. And so on. How many ways could you have these bits flipped?
    heheeh..i still stuck...i can't list it actually it will end until 1024 heehhe ( i refer to the answer) . so how to apply permutation here. what i can see it is a permutation which is involve semilar object..so nPn or nPr can't be use here...

    hope can u gauide me
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  9. #9
    A Plied Mathematician
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    Let's suppose you only had two bits. How many ways can you set the first bit? How many ways can you set the second bit? So, how many ways can you set both bits?
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  10. #10
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    Quote Originally Posted by Ackbeet View Post
    Let's suppose you only had two bits. How many ways can you set the first bit? How many ways can you set the second bit? So, how many ways can you set both bits?
    i had try but the ans is exceed the given ans in the book.
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  11. #11
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    I'm not interested in the book's answer. I'm interested in your answer. You need to think, here! How many ways can you set two bits?
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  12. #12
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    Quote Originally Posted by Ackbeet View Post
    I'm not interested in the book's answer. I'm interested in your answer. You need to think, here! How many ways can you set two bits?
    CASE 1
    ---------------------------------------------------------------------
    for 0000000001 , the number of way is=10!/(9!x1!)=10
    for 0000000010, the number of way is=10!/(9!x1!)=10
    .
    .
    for 1000000000, the number of way is=10!/(9!x1!)=10

    therefor the total the number of way is 10 x 10 = 100
    --------------------------------------------------------------------
    CASE 2

    for 0000000011 , the number of way is=10!/(8!x2!)=45
    therefore the total for case 2 is 45 x10 = 450
    ------------------------------------------------------------------
    CASE 3
    for 0000000111 , the number of way is=10!/(7!x3!)=120
    therefore the total for case 3 is 120 x10 = 1200

    if i add Case 1 to case 3 the ans already > ans in the text book

    huwahuwa...
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  13. #13
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    I'm afraid you're not thinking about the problem correctly. For each digit, you don't have 10 possibilities, you only have 2. Also, the possibilities don't add, they multiply. Just think about the two-bit case: you've got 00, 01, 10, 11. That's it. 4 possibilities. For 3 bits, you've got 000, 001, 010, 011, 100, 101, 110, 111. That's 8 possibilities. For 4 bits, you have 0000, 0001, 0010, 0011, 0100, 0101, 0110, 0111, 1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111; that's 16 possibilities. Do you see how this is working?
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  14. #14
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    Quote Originally Posted by Ackbeet View Post
    I'm afraid you're not thinking about the problem correctly. For each digit, you don't have 10 possibilities, you only have 2. Also, the possibilities don't add, they multiply. Just think about the two-bit case: you've got 00, 01, 10, 11. That's it. 4 possibilities. For 3 bits, you've got 000, 001, 010, 011, 100, 101, 110, 111. That's 8 possibilities. For 4 bits, you have 0000, 0001, 0010, 0011, 0100, 0101, 0110, 0111, 1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111; that's 16 possibilities. Do you see how this is working?
    tq for ur reply, totally im wrong..i will try it now
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  15. #15
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    update:
    i have try until bit no 5

    2 bit, they have 4 ways
    3 bit, they have 8 ways
    4 bit, they have 16 ways
    5 bit, they have 28 ways.

    i just stop and do the multipication until 5 bit to check the ans as below
    4x8x16x28>ans in the text book

    so how??
    any idea.. is it we need to add?
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