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Math Help - Combinatorics

  1. #1
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    Combinatorics

    This is the question. A computer has a password composed of 6 characters. The first 2 letters must be small letters, the rest can be either numbers or small letters.

    a) How many different passwords are possible?
    b) How many passwords are there that have no repeated characters?
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  2. #2
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    Quote Originally Posted by Keep View Post
    This is the question. A computer has a password composed of 6 characters. The first 2 letters must be small letters, the rest can be either numbers or small letters.

    a) How many different passwords are possible?
    b) How many passwords are there that have no repeated characters?
    Tell us what you do not understand about this question.
    At least, tell us how far you have gotten.
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  3. #3
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    Quote Originally Posted by Plato View Post
    Tell us what you do not understand about this question.
    At least, tell us how far you have gotten.
    First, the computer has a password of 6 characters. That means there is a possibility of up to 6! = 720 right? If no character is repeated, then we will still have 720 ways of writing the passowrd?
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  4. #4
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    Quote Originally Posted by Keep View Post
    First, the computer has a password of 6 characters. That means there is a possibility of up to 6! = 720 right?
    No. Because characters can be repeated. Only the first two must be alphabet.
    So the are 26^2 possiblities for the first two.
    Then the next four can be any one of 26+10=36.
    So what is the answer to part a)?
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  5. #5
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    So a is 36 + 676. Just to clarify on how you got 26. It is because of the number of ways in which the numbers can be entered which is 4! plus 2 for the characters or? Why did you raise it to the power of 2 then? And lastly, why did you add 10 to 26?
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  6. #6
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    Quote Originally Posted by Keep View Post
    So a is 36 + 676. Just to clarify on how you got 26. It is because of the number of ways in which the numbers can be entered which is 4!
    Absolutely not!
    Surely you know that there are 26 characters in the English alphabet.
    Do you have the basic ‘common body of knowledge’ to do this problem?
    For example: How many digits are there?
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  7. #7
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    Quote Originally Posted by Plato View Post
    Absolutely not!
    Surely you know that there are 26 characters in the English alphabet.
    Do you have the basic ‘common body of knowledge’ to do this problem?
    For example: How many digits are there?
    Thank you for your desire to help. Unfortunately, I haven't understood it very well because the resources with which we were provided was extremely limited. Nevertheless, I will try to understand it. First, the 26 is the number of characters in the English raised to 2 which is the first two characters which are alphabets. Then you add 26 to 10. 10 is the number of digits. This gives you 36 and you raise this to the power of 4. Then you add the  26^2. In other words you do 26^2 + 36^4
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  8. #8
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    Quote Originally Posted by Keep View Post
    . In other words you do 26^2 + 36^4
    Actually there are (26)^2(36)^4 as an answer for part (a).
    You explain that!

    Part (b), there are no repeats.
    The answer is (26)(25)\left(^{34}\mathcal{P}_4\right). WHY?
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  9. #9
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    Quote Originally Posted by Plato View Post
    Actually there are (26)^2(36)^4 as an answer for part (a).
    You explain that!
    This is because there are 26^2 ways for the first 2 characters and 36^4 for the remaining 4 characters which can either be digits or numbers (26+10)?



    Quote Originally Posted by Plato View Post
    Part (b), there are no repeats.
    The answer is (26)(25)\left(^{34}\mathcal{P}_4\right). WHY?
    This is because there are in total 26 characters and there are 26 ways for the first letter and 25 ways for the second letter then 34 ways for the remaining 4 characters: 10 digits and 24 letters?

    Any more explanation is appreciated, and thank you.
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  10. #10
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    Quote Originally Posted by Keep View Post
    Any more explanation is appreciated, and thank you.
    Think about it. The first two are alpha.
    The next four are either alpha or digit.
    So the last four can be chosen in (34)(33)(32)(31) ways.
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  11. #11
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    Ok, I understood it now. Thank you once again for your patience and help! Understanding is always better than getting just an answer.
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  12. #12
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    Quote Originally Posted by Plato View Post
    \left(^{34}\mathcal{P}_4\right).

    Hi,

    I'm just wondering what that means - I don't think I've come across that notation before.

    Is it just \left(^{n}\mathcal{P}_i\right) = n(n-1) \ldots (n-i+1)?

    Thanks.
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