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Math Help - The number of possible PIN numbers

  1. #1
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    The number of possible PIN numbers

    This seems like a pretty straightforward question but I feel like I'm missing something.

    Suppose I wanted to construct a pin number that contains 4 letters of the alphabet and 3 numbers (0 to 9) in no particular order, and I can use the same letter/number more than once.

    My pin will be 7 letters/numbers in length.

    I have:

    (26)(26)(26)(26)(10)(10)(10)

    = 456 976 000

    It feels like I'm missing something or is the question really this straightforward?

    Any help is appreciated!
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  2. #2
    MHF Contributor harish21's Avatar
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    That is correct!

    Had it been without repeating, you would have had: 26.25.24.23.10.9.8
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  3. #3
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    So really, this is essentially the same thing if I wanted my pin number to start with 4 letters and then followed by 3 numbers? (I know this is a stupid question)

    I don't have to divide out anything at all?
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  4. #4
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    Quote Originally Posted by DarK View Post
    So really, this is essentially the same thing if I wanted my pin number to start with 4 letters and then followed by 3 numbers? (I know this is a stupid question)

    I don't have to divide out anything at all?
    Your calculation counted the amount of pin numbers in the sequence LLLLDDD, where L is a letter and D is a digit,
    allowing repetition.

    That's fine if no shuffling of the letters among the digits is allowed.
    (If you meant that the 4 letters are in direct sequence "in no particular order" and same for the digits).
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  5. #5
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    So what if I could shuffle the letters and numbers in any way I want:

    ie:

    LDDLDLL
    or
    DLDLLDL
    or
    DDLLLDL

    and so on

    Would I have to account for this or would it just be the same calculation as my first post?

    Thank you.
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  6. #6
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    Quote Originally Posted by DarK View Post
    So what if I could shuffle the letters and numbers in any way I want: ie: LDDLDLL or DLDLLDL
    \displaystyle \binom{7}{4}\left( {26^4 } \right)\left( {10^3 } \right)
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  7. #7
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    Quote Originally Posted by Plato View Post
    \displaystyle \binom{7}{4}\left( {26^4 } \right)\left( {10^3 } \right)
    Could somebody clarify, why are we multiplying by 7 choose 4?
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  8. #8
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    Quote Originally Posted by DarK View Post
    Could somebody clarify, why are we multiplying by 7 choose 4?

    \binom{7}{4}=\binom{7}{3}

    We choose 4 of the 7 positions for the 4 letters.
    That automatically leaves us with 3 positions for the 3 digits.

    Now you have 26^4 alternatives for the letters and 10^3 alternatives for the digits in that particular sequence.

    the sequence could be LDLDLDL

    Now choose a new sequence and you have the same 26^4 alternatives for the letters and 10^3 for the digits.

    The number of such "letter-digit" sequences is \binom{7}{4}=\binom{7}{3}

    as you need to choose 4 of the 7 positions for letters
    or instead choose 3 of the 7 positions for the digits (same thing).

    Hence you need to count the number of ways this can be done.
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  9. #9
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    So it looks like the final answer can be either:

    \binom{7}{4} 26^4 10^3

    or

    \binom{7}{3} 26^4 10^3

    where the letters and numbers can take on any ordering.
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