Digits Problem: Would you look at the solution for the following problem and comment on whether it is correct? Thanks. Problem appears first in thread followed by proposed solution.

Question

directrix · Answer

If the six digits 1, 2, 3, 5, 5, and 8 are randomly arranged into a six digit positive integer (number), what is the probability that the integer (number) is divisible by 15?  Write your answer as a fraction in lowest terms.

directrix · Answer

To be divisible by 15 a number has to be divisible by 3 and by 5. To be divisible by 5 the number must end in a 0 or a 5.  However, divisibility by 3 is not as simple - the rule is to add up all of the digits of the number and if the sum is  a multiple of 3 then the original number is also
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The sum of these digits 1, 2, 3, 5, 5, and 8 is 24 which is a multiple of 3.

Of the six digits, two are the number 5 and there are no 0s.

For the last digit, there are two choices: 5 or 5. After using one of these two choices for the last digit, there are five choices left for the remaining 5 digits: 1,2,3,5,8.

There are 5*4*3*2*1*2 = 240 ways to create a number divisible by 15 from the digits.
The two fives are indistinguishable. Hence, there are 240/2 = 120 ways to write distinct numbers that adhere to the requirements.

120 will be the numerator of the probabiity fraction.

For the six digits, there are: 6*5*4*3*2*1 = 720 possible numbers. Because of the two fives, there are 360 distinct possible numbers.

P(the integer (number) is divisible by 15) = 120/360 = 1/3.

kinggeorge · Answer

Looks correct and makes sense to me.

directrix · Answer

If I don't divide out the repetitions for the numerator and denominator, the probability answer is the same. I thought about not mentioning it in the explanation for the student. But, somehow, I am thinking the duplications have to matter but not here, it seems. @PaxPolaris

paxpolaris · Answer

you're right it's still 1/3

directrix · Answer

Then, I was looking at the following problem:

 How many different messages can be represented by sequences of length 3 that use two dashes and one dot?

and thinking 3*2*1 = 6 but because of the two indistinguishable dashes, then the answer should be 3.

Does anybody agree?

kinggeorge · Answer

I agree. Possibilities are 
.--
-.-
--.

anonymous · Answer

That is corect.
Even the probability thing.
Any repitions - You divide their interarrangeent sit doesnt matter. 
So if n similar things are thee in arranements you divide by n!.