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Probability of combination lock

  1. May 11, 2013 #1
    1. The problem statement, all variables and given/known data
    a standard combination lock consists of a password which is 3 distinct numbers between 00 and 59. a) compute the number of valid passwords
    b) compute the probability that the password is a palindrome. eg. 03 - 22 - 30 is a palindrome (same backward as it is forward)
    c) the product of the three numbers is a factor of 5
    d) the password forms an arithmetic sequence with difference k>0



    2. The attempt at a solution

    a) Valid passwords = P(60,3) (since none can be repeated)
    b) This is possible only if the first number is the first digit is 0 to 5 - 6 possilbities, however, we cannot use the same number twice - cant use 00 in the first and last place so only 5
    Suppose the first digit is zero such as 00, the middle number must be a number whose units and tens place are identical such as 11, 22, 33 ... 55- 6 possibilities.
    THe last number is decided by the first so 1 there
    Each case has 30 possiblities and since we can only allow first digit 0, 1, 2, 3, 4, 5 - 6 so 180 possibilities

    c) We can do it by compliment - calculate the number of possiblities whose numbers are NOT a factor of 5 (exclude 00, 05, ... 55 - 12 possiblities is total) leaving 48 possibilities
    the total number of possibilities is P(48,3)

    d) I thought this question was really hard
    I did it by cases
    When the first number is 0 to 9 , we can allow up 1 to 30 in the second digit. The 30 gets reduced by 1 for each number we add to the first so 10 allows 30 in the second, 11 allows 29 in the second and so on. 10 x 30!
    first number 10 to 19, second number allows 15 numbers as it allows 11 through 25 in the second place 10 x 15!
    first number 20 to 29, second number allows 15 as well?

    According to this, the second number allows 15 as long as the number is 10 to 39 so
    10 x 15! x 3
    if the first number is 40 to 49, then 10 x 10! x 1
    if the first number is 50 to 59, then 10 x 10! x 1

    When we add all of these up we get the final answer and that is the total number of possibilites with sequences.

    Is this all correct? Especially the last one?

    Thanks
     
  2. jcsd
  3. May 11, 2013 #2

    haruspex

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    I confirm your answers for a, b and c, though I didn't follow your reasoning for b.
    For d, you seem to be reading it as an arithmetic sequence of 6 digits. I read it as a sequence of 3 two-digit numbers, which is much easier.
     
  4. May 11, 2013 #3
    How many factors of 5 are there?

    Or did you mean "multiple"?
     
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