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.
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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
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.