Probability of combination lock

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SUMMARY

The discussion focuses on calculating the number of valid passwords for a standard combination lock consisting of three distinct numbers between 00 and 59. The valid passwords are determined using permutations, specifically P(60,3), resulting in 180 possibilities for palindromic combinations. The probability that the product of the three numbers is a factor of 5 is computed by excluding combinations that include multiples of 5, leading to P(48,3). The arithmetic sequence question is clarified as involving three two-digit numbers, simplifying the calculations significantly.

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  • Understanding of permutations, specifically P(n, r)
  • Knowledge of palindromic sequences in number theory
  • Familiarity with factors and multiples, particularly in relation to the number 5
  • Basic concepts of arithmetic sequences
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  • Study combinatorial mathematics, focusing on permutations and combinations
  • Learn about palindromic numbers and their properties
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stunner5000pt
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Homework Statement


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 - can't 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
 
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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.
 
stunner5000pt said:
c) the product of the three numbers is a factor of 5
How many factors of 5 are there?

Or did you mean "multiple"?
 

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