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Math Help - permutations (password problem)

  1. #1
    baz
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    permutations (password problem)

    Can any check out this one?

    A password for the computer is a string of digits
    0
    , 1, 2, 3, 4, 5, 6, 7, 8, 9
    and/or letters
    a,b,c,. . . ,x,y,z
    (there are 26 different letters). The password must obey the following
    rules:
    (a) the length of the password must be exactly 8 characters;
    (b) the password must contain at least one digit, and no more than
    three digits.

    What is the number of different passwords you can form?

    I think number of different passwrds are
    10x26! + 10x10x25! +10x10x10x24!
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  2. #2
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    Baz, may I ask if you understand any of these problems?
    You seem to be gussing at answers.

    This time I shall give you the answer as it might appear in the 'back of the book'

    \sum\limits_{k = 1}^3 {\binom{8}{k}\left( {10^k } \right)\left( {26^{8 - k} } \right)}

    Can you explain that to us?
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  3. #3
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    Quote Originally Posted by baz View Post
    Can any check out this one?

    A password for the computer is a string of digits
    0
    , 1, 2, 3, 4, 5, 6, 7, 8, 9
    and/or letters
    a,b,c,. . . ,x,y,z
    (there are 26 different letters). The password must obey the following
    rules:
    (a) the length of the password must be exactly 8 characters;
    (b) the password must contain at least one digit, and no more than
    three digits.

    What is the number of different passwords you can form?

    I think number of different passwrds are
    10x26! + 10x10x25! +10x10x10x24!
    3 things to note, baz.....

    1. Do you realise you are making a password violating rule (a) ?

    2. In forming your 27-character password, do you realise that you are only arranging the letters ?

    3. Do you also realise that you are allowing the digits to be used multiple times but not the letters ?
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  4. #4
    baz
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    Quote Originally Posted by Plato View Post
    Baz, may I ask if you understand any of these problems?
    You seem to be gussing at answers.

    This time I shall give you the answer as it might appear in the 'back of the book'

    \sum\limits_{k = 1}^3 {\binom{8}{k}\left( {10^k } \right)\left( {26^{8 - k} } \right)}

    Can you explain that to us?
    Plato you have given the answer for the case where repitition is allowed.
    Case1 (1 digit allowed)
    10 ways of choosing one (10 digits )of the 8 spots , 26^7 ways of choosing remaining spots , (8 binom 1) ways of choosing 8 taking 1 at a time.
    by product rule all the above outcomes are multiplied.

    Similarly we will workout ways for remaining two cases and them .
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