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anonymous

  • one year ago

HELP PLEASE FAN, MEDAL AND TESTIMONIAL Idi is creating a password for a website that has some strict requirements. The password must be 8 characters. Numbers and letters may be used, but may not be repeated. How many diffferent passwords are possible? How many are possible if the following restrictions are enforced; The password must feature both numbers and letters? The password must start with a letter? The password must start with a letter and end with a number?

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  1. anonymous
    • one year ago
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    Do you know?

  2. anonymous
    • one year ago
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    permutation and combination unit. I know it's tricky.

  3. anonymous
    • one year ago
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    Guys I know this is super difficult... Would really appreciate your insight

  4. Vocaloid
    • one year ago
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    well, let's start off with the first question "how many different passwords are possible?" we can choose any of the 10 digits or any of the 26 letters (the question doesn't specify whether the password is case sensitive or not, so I'm going to go ahead and use 26). this gives us a possible of 10+26 = 36 possibilities for each of the 8 characters in the password so, for the first character we have 36 choices for the second character, we have 35 choices since we already used one for the third character, we have 34 choices, and so on, until we get to the last character total number of possibilities = (36)*(35)*(34)*(33)*(32)*(31)*(30)*(29)

  5. anonymous
    • one year ago
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    Right. So that's a typical permutation without restrictions.

  6. anonymous
    • one year ago
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    So the answer would be 36P36

  7. Vocaloid
    • one year ago
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    well, 36P8, but you have the right idea

  8. anonymous
    • one year ago
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    Oh yeah you are right.

  9. Vocaloid
    • one year ago
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    ok, we can move on to the second question now "The password must feature both numbers and letters?"

  10. anonymous
    • one year ago
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    I would think that it has to do with 10 and 26?

  11. anonymous
    • one year ago
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    I need to take into account 36 options AND somehow play with 10 and 26 because password has to feature both

  12. Vocaloid
    • one year ago
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    right, we want to find out A. how many passwords only use numbers and B. how many passwords only use letters then, we want to figure out 36P8 - A - B

  13. anonymous
    • one year ago
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    Oh, so that's left would be the answer.

  14. Vocaloid
    • one year ago
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    right!

  15. anonymous
    • one year ago
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    Let me think about A.,,,

  16. anonymous
    • one year ago
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    26P8?

  17. anonymous
    • one year ago
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    10P*?

  18. anonymous
    • one year ago
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    10P8

  19. Vocaloid
    • one year ago
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    right, numbers only would be 10P8 and letters only would be 26P8 ready to keep going?

  20. anonymous
    • one year ago
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    Yes! So 36P8-(26P8+10P8)=possible combinations including both letters and numbers.

  21. anonymous
    • one year ago
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    Wow you are so good

  22. Vocaloid
    • one year ago
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    haha, thanks ~ next question is "The password must start with a letter" so, for the first character, we have 26 possibilities since we can only use letters for the second character, we have 36-1=35 possibilities, since we already used one of the letters then 34, then 33, and so on until we get to the 8th character

  23. anonymous
    • one year ago
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    35P8 IT IS!!!!

  24. anonymous
    • one year ago
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    What a concise explanation

  25. anonymous
    • one year ago
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    :):)

  26. Vocaloid
    • one year ago
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    well, it would actually be 26*[35P7]

  27. anonymous
    • one year ago
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    Accounting all the possibilities for a letter right?

  28. anonymous
    • one year ago
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    Yes I get it.

  29. Vocaloid
    • one year ago
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    alright, last one is the trickiest, it might take me a minute to get the answer "The password must start with a letter and end with a number?"

  30. anonymous
    • one year ago
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    I think 26*(35P7)*10something

  31. anonymous
    • one year ago
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    Since I'd have to account for the numbers used until the last digit.

  32. anonymous
    • one year ago
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    But knowing that is probably the trickiest.

  33. Vocaloid
    • one year ago
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    26*(35P6)*10 would be my best guess

  34. Vocaloid
    • one year ago
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    however...

  35. anonymous
    • one year ago
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    Would that number ever be repeated?

  36. anonymous
    • one year ago
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    Like say some number was already used before hand

  37. Vocaloid
    • one year ago
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    yes, that's what I was thinking too

  38. Vocaloid
    • one year ago
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    we have to account for the first and last character first, before we tackle the middle

  39. anonymous
    • one year ago
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    ok

  40. anonymous
    • one year ago
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    -n?

  41. Vocaloid
    • one year ago
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    so, we have 26 possibilities for the first character and 10 possibilities for the last character after those two characters are used up, we have 36-2 = 34 characters left for the 6 digits in the middle, giving us (26)*(34P6)*(10)

  42. anonymous
    • one year ago
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    That looks about the closest to being right? I don't quite get how we are supposed to know if numbers may have been used in the (34P6) part in which case the last digit option would decrease instead of full 10?

  43. anonymous
    • one year ago
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    But within our limits of intuition yours seems to be the rightest.

  44. anonymous
    • one year ago
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    (26)*(34P6)(10-N)?

  45. anonymous
    • one year ago
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    n being how many numbers already used until then

  46. Vocaloid
    • one year ago
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    right, I was thinking the same thing, it would be best to double-check with another, though @ganeshie8 @satellite76

  47. anonymous
    • one year ago
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    so you summoned them?

  48. anonymous
    • one year ago
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    You are so nice:)

  49. anonymous
    • one year ago
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    Are you a math student in uni?

  50. Vocaloid
    • one year ago
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    yeah, I'm only a sophomore though so I haven't done anything past Calculus/Linear Algebra though

  51. anonymous
    • one year ago
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    Nice! Is life full of discovery for you now?

  52. anonymous
    • one year ago
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    I am stuck for a month or two in high school:(

  53. anonymous
    • one year ago
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    Hi everyone Vocaloid and I had arrived at the answer 26*(34P6)*(10-n) where n is the number of numbers used in making 8 digit password composed of letters and numbers. I would really appreciate your insights as well

  54. anonymous
    • one year ago
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    I think your last answer is right Vocaloid:)

  55. anonymous
    • one year ago
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    No other way can it be expressed

  56. Vocaloid
    • one year ago
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    Right, that's my best guess. I'm still a bit uncertain about the whole "how do we account for situations where numbers have already been used" but eh

  57. anonymous
    • one year ago
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    that's why I think it's n

  58. anonymous
    • one year ago
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    I don't think there is any degree of certainty until the last digit

  59. anonymous
    • one year ago
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    because if all 6 digits be numbers, then the last one would be 26*(34PN)*(10-6)

  60. anonymous
    • one year ago
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    26*(34P6)*(10-6)

  61. anonymous
    • one year ago
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    So n be the determinant for the possiblities

  62. ganeshie8
    • one year ago
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    26*(34P6)*(10) is right for last part

  63. anonymous
    • one year ago
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    not n?

  64. ganeshie8
    • one year ago
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    nope

  65. anonymous
    • one year ago
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    What if the numbers were used in until the last digit? That would limit how many numbers can be picked.

  66. ganeshie8
    • one year ago
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    good question, think of it this way : first pick first and last characters, then that ambiguity wont be there.

  67. anonymous
    • one year ago
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    Oh

  68. anonymous
    • one year ago
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    you are right .

  69. ganeshie8
    • one year ago
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    In the start, we have 26+10 = 36 characters to choose from. But we want the first character to be a letter, so there are 26 ways to choose first character : |dw:1437017702350:dw|

  70. anonymous
    • one year ago
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    Yes

  71. anonymous
    • one year ago
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    |dw:1437017772324:dw|

  72. ganeshie8
    • one year ago
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    After that, we choose the last character instead of the second. We have 35 characters to pick from, but we want the last character to be a number. so we have 10 ways to choose a number for last character : |dw:1437017797071:dw|

  73. ganeshie8
    • one year ago
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    I see you get it :)

  74. anonymous
    • one year ago
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    So happy now:) Thanks so much Ganeshie8!

  75. anonymous
    • one year ago
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    Thanks everyone!

  76. ganeshie8
    • one year ago
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    np, thank Vocal

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