Probability of combination lock

AI Thread Summary
The discussion revolves around calculating probabilities and valid combinations for a standard combination lock with distinct numbers between 00 and 59. For part a, the number of valid passwords is determined using permutations, yielding P(60,3). Part b explores the conditions for a palindrome, concluding there are 180 valid combinations. Part c involves calculating combinations where the product of the numbers is a factor of 5, leading to 48 possibilities after excluding certain combinations. Part d discusses the arithmetic sequence requirement, with some confusion regarding the interpretation of the sequence length and the calculation method.
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"?
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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