Question 1034565: A six digit number is formed using the digits 1, 2, 3, 5, 5, and 8, each exactly once. What is the probability that the resulting number is divisible by 15?
Answer by robertb(5830)   (Show Source):
You can put this solution on YOUR website! For the six-digit number to be divisible by 15, it has to be divisible by both 3 and 5.
It is easy to see that any arrangement of the six digits will always be divisible by 3, because the sum of the digits itself is divisible by 3. (Sum of the digits is 24.) We just have to make sure that the number is divisible by 5.
To make the number divisible by 5, the six-digit number has to end in 5.
Fix the position of one of the fives at the ones place of the number. This means we only have to arrange the digits 1, 2, 3, 5, and 8 before the fixed 5. This can be done in 5! = 120 ways, being distinct.
Now, without any restrictions, the six-digits can be arranged in  ways.
Therefore the probability that the resulting number is divisible by 15 is  .
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