Question 1093789: The total no. Of no. Less than 1000 and divisible by 5 formed with 0,1,2.......9 such that each digit does not occur more than once in each no.is
Answer by greenestamps(13198)   (Show Source):
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If the number is divisible by 5, then the last digit is either 0 or 5. So you have two cases to analyze.
If the last digit is 0, then there are 9 choices for one of the other digits and then 8 choices for the third digit. By the fundamental counting principle, the number of numbers less than 1000, divisible by 5, ending in 0, and containing 3 different digits is 9*8 = 72.
If the last digit is 5, there are still 9 choices for one of the other digits. That is because the problem said "numbers less than 1000", not "3-digit numbers" -- so the first digit can be 0. And then again there are 8 choices for the last digit, making again 9*8=72 numbers less than 1000, divisible by 5, ending in 5, and containing 3 different digits.
So there are a total of 72+72=144 numbers that are less than 1000, are divisible by 5, and contain 3 different digits.
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